mathamatics xam idea

30. Draw a circle of radius 4 cm. Take two points P and Q on one of its extended diameter each at a
distance of 8 cm from its centre. Draw tangents to the circle from these two points P and Q.

31. Draw a circle with the help of any circular object. Take a point outside the circle and construct
the pair of tangents from this point to the circle.

32. Given a rhombus ABCD in which AB = 4 cm and ∠ABC = 60°, divide it into two triangles say,
DABC and DADC. Construct the triangle AB'C' similar to DABC with scale 2 . Draw a line
factor 3

segment C'D' parallel to CD where D' lies on AD. Is AB'C'D' a rhombus? Give reasons.

33. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and

taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the

centre of the other circle. [CBSE Delhi 2015]

34. Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a

point on the larger circle. Also measure its length. [CBSE (Delhi) 2016]

35. Draw a DABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then construct another triangle

whose sides are 3 of the corresponding sides of DABC. [CBSE (F) 2016]
5

Answers

1. (i) (c) (ii) (b) (iii) (a) (iv) (d) (v) (b)

2. (a+b) 3. P10 4. Join B5C 5. Join B5C 6. 9
7. 100° 8. Yes 14. Do yourself
19. 4.5 cm (approx)

23. 4.5 cm approx 29. Yes

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (6 × 1 = 6)

(i) To divide a line segment AB in the ratio p:q (p. q are positive integers), draw a ray AX so

that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that

the minimum number of these points is [NCERT Exemplar]

(a) greater of p and q (b) p + q

(c) p + q – 1 (d) pq

(ii) To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°,

it is required to draw tangents at the end points of those two radii of the circle, the angle

between them is [NCERT Exemplar]

(a) 105° (b) 70° (c) 140° (d) 145°

(iii) If in the construction of triangle the ratio of side is given in improper fraction form then

the new triangle will be [NCERT Exemplar]

(a) bigger (b) smaller (c) equal (d) can’t say

Constructions 243

(iv) To construct a triangle similar to a given ∆ABC with its sides 3 of the corresponding sides
7
of ∆ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite

side of A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and

next step is to join [NCERT Exemplar]

(a) B10 to C (b) B3 to C (c) B7 to C (d) B4 to C

(v) To draw a pair of tangents to a circle which are inclined to each other at an angle of 40°,
it is required to draw tangents at the end points of those two radii of the circle, the angle
between which is

(a) 140° (b) 135° (c) 100° (d) 120°

(vi) If the centre of a circle is not given, it can be located by [NCERT Exemplar]
(a) finding perpendicular bisectors of parallel chords
(b) finding perpendicular bisectors of non parallel chords
(c) Either (a) or (b)
(d) None of these

2. Fill in the blanks. (4 × 1 = 4)
(i) A plane closed figure bounded by three straight lines is ________________.

(ii) Number of points common to a circle and one of its tangents is ________________.

(iii) Only ________________ tangent can be drawn from a point on the circle.

(iv) We can construct a triangle similar to a given triangle as per given ________________ which
may be less than 1 or greater than 1.

Section B

QQ Solve the following questions. (5 × 3 = 15)

3. Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose

sides are 3 times the corresponding sides of the given triangle. [CB SE (AI) 2014]
5

4. Construct a triangle ABC, in which AB = 5 cm, BC = 6 cm and AC = 7 cm. Then construct

another triangle whose sides are 3 times the corresponding sides of DABC. [CBSE (F) 2014]
5

5. Draw a line segment of length 7.4 cm and divide it in the ratio 4 : 7. Measure the two parts.

6. Construct an equilateral triangle of side 4.8 cm and then another triangle whose sides are 7 of
4
the corresponding sides of the first triangle.

7. Draw a right triangle in which the sides (other than hypotenuse) are 8 cm and 6 cm. Then

construct another triangle whose sides are 3 times the (corresponding) sides of given triangle.
4 [CBSE Delhi 2017 (C)]

244 Xam idea Mathematics–X

QQ Solve the following questions. (3 × 5 = 15)

8. Two line segments AB and AC include an angle of 60°, where AB = 5 cm and AC = 7 cm. Locate
3 1
points P and Q on AB and AC, respectively such that AP = 4 AB and AQ = 4 AC. Join P and

Q and measure the length PQ. [NCERT Exemplar]

9. Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle, construct

the pair of tangents to the other. Measure the length of a tangent and verify it by actual

calculation. [NCERT Exemplar]

10. Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and

3 cm. Now construct another triangle whose sides are 3 times the corresponding sides of the
given triangle. 5 [CBSE Delhi 2017]

Answers

1. (i) (b) (ii) (d) (iii) (a) (iv) (c) (v) (a)

(vi) (b)

2. (i) triangle (ii) one (iii) one (iv) scale factor

5. 2.7 cm and 4.7 cm 8. 3.25 cm 9. 4 cm

zzz

Constructions 245

10 Introduction to
Trigonometry
BASIC CONCEPTS – A FLOW CHART

246 Xam idea Mathematics–X

MORE POINTS TO REMEMBER

 sin q is an abbreviation for “sine of angle q”. It is not the product of sin and q. Therefore, we
cannot do like this

   sin 60° = 60° = 2°
sin 30° 30°

 It holds for other trigonometric ratios also.

 Trigonometric ratios are defined for an acute angle 0° ≤ q ≤ 90°.

 The trigonometric ratios are same for the same angle.

For example, If in Fig. 10.1,

∠ABC = ∠A'B'C' = q, then

sin q = AC = A'C'
AB A'B'

Proof: Obviously, DABC ~ DA'B'C' as

∠B = ∠B' and ∠C = ∠C'

Therefore,

AB = AC = BC & AB = AC Fig. 10.1
A'B' A'C' B'C' A'B' A'C'

A'C' = AC = sin T
A'B' AB

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. The value of (sin 30° + cos 30°) – (sin 60° + cos 60°) is [NCERT Exemplar]

(a) – 1 (b) 0 (c) 1 (d) 2
[NCERT Exemplar]
2. The value of tan 30° is
cot 60°

(a) 1 (b) 1 (c) 3 (d) 1
2 3

3. The value of (sin 45° + cos 45°) is [NCERT Exemplar]

(a) 1 (b) 2 (c) 3 (d) 1
2 2

4. If cos A +cos2 A = 1, then sin2 A + sin 4 A is

(a) – 1 (b) 0 (c) 1 (d) 2

5. If 6cotθ + 2cosecθ = cotθ + 5cosecθ, then cosθ is

(a) 4 (b) 5 (c) 3 (d) 5
5 3 5 4

6. The value of 2 tan 30° is equal to
1 – tan2 30°

(a) cos 60° (b) sin 60° (c) tan 60° (d) sin 30°

7. 9 sec2 A – 9tan2 A is equal to

(a) 1 (b) 9 (c) 8 (d) 0

Introduction to Trigonometry 247

8. (1 + tanθ + secθ) (1 + cotθ – cosecθ) is equal to

(a) 0 (b) 1 (c) 2 (d) – 1

9. If tan A = 3 , then the value of cos A is
2

(a) 3 (b) 2 (c) 2 (d) 13
13 13 3 2

10. If sin A = 1 , then the value of cot A is [NCERT Exemplar]
2
1 3
(a) 3 (b) 3 (c) 2 (d) 1

11. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

[NCERT Exemplar]
3
(a) – 1 (b) 0 (c) 1 (d) 2

12. Given that sin i = a , then cos θ is equal to [NCERT Exemplar]
b
b b b2 – a2 a
(a) b2 – a2 (b) a (c) b (d) b2 – a2

13. If cos (α + β) = 0, then sin (α – β) can be reduced to [NCERT Exemplar]

(a) cos β (b) cos 2β (c) sin α (d) sin 2α

14. The value of (tan1° tan2° tan3°... tan89°) is 1 [NCERT Exemplar]
2
(a) 0 (b) 1 (c) 2 (d)

15. If cos 9α = sinα and 9α <90, then the value of tan 5α is [NCERT Exemplar]

(a) 1 (b) 3 (c) 1 (d) 0
3

16. If ∆ABC is right angled at C, then the value of cos (A + B) is [NCERT Exemplar]

(a) 0 (b) 1 (c) 1 (d) 3
2 2
1 1
17. Given that sin a = 2 and cos b = 2 , then the value of (α +β) is [NCERT Exemplar]

(a) 0° (b) 30° (c) 60° (d) 90°

18. The value of the expression sin 60° is
cos 30°

(a) 3 (b) 1 (c) 1 (d) 2
2 2

19. If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is [NCERT Exemplar]

(a) 1 (b) 1 (c) 2 (d) 3
2

20. If 4 tanθ = 3, then 4 sin i – cos i is equal to [NCERT Exemplar]
d 4 sin i + cos i n
2 1 1 3
(a) 3 (b) 3 (c) 2 (d) 4

Answers

1. (b) 2. (d) 3. (b) 4. (c) 5. (c) 6. (c)
7. (b) 8. (c) 9. (b) 10. (a) 11. (b) 12. (c)
13. (b) 14. (b) 15. (c) 16. (a) 17. (d) 18. (c)

19. (a) 20. (c)

248 Xam idea Mathematics–X

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s). 5. cosθ
11. less
1. Reciprocal of sin θ is ______________ .

2. Sum of ______________ of sine and cosine of an angle is one.

3. Triangle in which we study trigonometric ratios is called______________ .

4. Cosine of 90° is ______________.

5. sine of (90 – θ) is ______________ .

6. Maximum value for sine of any angle is ______________ .

7. tcaont 47° = ______________ .
43°

8. The value of the expression [cos223° – sin267°] is ______________ to zero.

9. The value of the expression (sin 80° – cos 80°) is ______________ than zero.
10. The expression (1 – cos2 i) sec2 i = ______________ .

11. The value of sinq + cosq is always ______________ than or equal to 1.

12. cot q is not defined for q = ______________ .
13. Value of sinq ______________ as q decreases
14. Value of cosq decreases as q______________ .
15. sinq = cosq for q = ______________ .

Answers

1. cosecq 2. square 3. right triangle 4. zero
6. 1 10. tanq
12. 0° 7. 1 8. equal 9. greater

13. decreases 14. increases 15. 45°

Very Short Answer Questions [1 mark]

1. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. [NCERT]
Sol. sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°
2. Evaluate:

(i) sin 18° (ii) cos 48° – sin 42° [NCERT, CBSE 2019 (30/5/1)]
cos 72°

Sol. (i) sin18° = sin (90° − 72°) = cos 72° =1
cos 72° cos 72° cos 72°

(ii) cos 48° – sin 42°  = cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

3. In DABC right angled at C, find the value of cos (A + B).

Sol. In Fig. 10.2

 ∠C = 90°

\ ∠A + ∠B = 90° (By angle sum property)

Hence, cos (A + B) = cos 90° = 0.

Fig. 10.2

Introduction to Trigonometry 249

4. Can the value of the expression (cos 80° – sin 80°) be negative? Justify your answer.

Sol. Yes, for q > 45°, sin q > cos q, so cos 80° – sin 80° has a negative value.
5. If sin A + sin2 A= 1, then show that cos2 A + cos4 A = 1.

Sol. Given: sin A + sin2 A = 1  ⇒ sin A = 1 – sin2 A = cos2 A
\ LHS = cos2 A + cos4 A = sin A + sin2 A = 1 = RHS.

6. If sinx + cosy = 1; x = 30° and y is an acute angle, find the value of y. [CBSE 2019 (30/5/1)]

Sol. We have, sin x + cos y = 1

⇒ sin30° + cos y = 1

⇒ 1 + cos y =1 & cos y =1 – 1 = 1 ⇒ y = 60°
7. If sin 2 2 2 = 2× 9 = 18
Sol. 2cot2 = 1 , then find the value of 2cot2 q + 2.
q 3 2 2
q +2= 2(cot2 q + 1) = 2cosec2 q = sin2 θ  1 2
=

 3 

8. If sec2 q(1 + sin q) (1 – sin q) = k, then find the value of k.

Sol. we have, sec2 q(1 + sin q)(1 – sin q) = sec2 q(1 – sin2 q) ( (a + b)(a – b) = a2 – b2)

= sec2 q.cos2 q = 1 ( cos2 q + sin2 q = 1)

\ k = 1

9. Write the acute angle q satisfying 3 sin q = cos q.

Sol. We have, 3 sin θ = cos θ

⇒ sin θ = 1 ⇒ tan θ = 1 ⇒ θ = 30°
cos θ 3 3

10. If A + B = 90° and tan A = 3 , what is cot B?
4

Sol. We have, cot B = cot (90° – A) ( A + B = 90°)

= tan A = 3 . ( cot (90° – q) = tan q)
4

Short Answer Questions-I [2 marks]

1. Evaluate cos 48° cos 42° – sin 48° sin 42°.

Sol. cos 48° cos 42° – sin 48° sin 42° = cos (90° – 42°) cos (90° – 48°) – sin 48° sin 42°

= sin 42° sin 48° – sin 48° sin 42°   ( cos (90° – q) = sin q)

= 0

2. Find the value of 3sin2 20° – 2tan2 45° + 3sin2 70°.

Sol. 3sin2 20° – 2tan2 45° + 3sin2 70°

= 3sin2 (90° – 70°) – 2(1)2 + 3sin2 70° ( tan 45° = 1)

= 3 cos2 70° – 2 + 3 sin2 70° ( sin (90° – q) = cos q)
= 3 (sin2 70° + cos2 70°) – 2

= 3 × 1 – 2 = 3 – 2 = 1. ( sin2 q + cos2 q = 1)

3. If sin2 A = 2 sin A then find the value of A.

Sol. sin2 A = 2 sin A

⇒ sin2 A – 2 sin A = 0 ⇒  sin A (sin A – 2) = 0

⇒ either sin A = 0 or sin A – 2 = 0.

⇒ A = 0° (sin A = 2, which is not possible)

\ Value of ∠A = 0°

250 Xam idea Mathematics–X

4. Find maximum value of 1 , 0° ≤ q ≤ 90°.
sec i
1
Sol. sec i , (0° ≤ q ≤ 90°) (Given)

 sec q is in the denominator. 1 .
\ The min. value of sec q will return max. value for sec i
But the min. value of sec q is sec 0° = 1.

Hence, the max. value of 1 = 1 =1.
sec 0° 1

5. Given that sin q = a , find the value of tan q.
b
a
Sol. sin q = b

⇒ cos q= 1 – sin2 q= 1 − ab22= b2 − a2= b2 − a2
b2 b

tan i = sin i = a/b = a
cos i b2 – a2 b2– a2

b

6. If sin q = cos q, then find the value of 2 tan q + cos2 q.

Sol. sin q = cos q (Given)

It means value of q = 45°

Now, 2 tan q + cos2 q = 2 tan 45° + cos2 45°

= 2×1+e 1 2 = 2 + 1 = 4+1 = 5 ea tan 45° =1, cos 45° = 1
2 2 2 2 2o
o

7. If sin (x – 20)° = cos (3x – 10)°, then find the value of x.
Sol. sin (x – 20)° = cos (3x – 10)°
⇒ cos [90° – (x – 20)°] = cos (3x – 10)°
By comparing the coefficient

90° – x° + 20° = 3x° – 10° ⇒ 110° + 10° = 3x° + x°

120° = 4x° ⇒ x° = 120° = 30°
4

8. If sin2A = 1 tan2 45°, where A is an acute angle, then find the value of A.
2

Sol. sin2A = 1 tan2 45° ⇒ sin2A = 1 (1)2 ( tan 45° = 1)
2 2

⇒ sin2 A = 1 ⇒ sin A = 1
2 2

Hence, ∠A = 45°

9. If x = a cos q, y = b sin q, then find the value of b2x2 + a2y2 – a2b2.

Sol. Given x = acos q, y = b sin q
b2x2 + a2y2 – a2b2 = b2(acos q)2 + a2(b sin q)2 – a2b2

= a2b2 cos2 q + a2b2 sin2 q – a2b2 = a2b2 (sin2 q + cos2 q) – a2b2

= a2b2 – a2b2 = 0 ( sin2 q + cos2 q = 1)

Introduction to Trigonometry 251

10. If tan A = cot B, prove that A + B = 90°. [NCERT]

Sol. We have

tan A = cot B ⇒ tan A = tan (90° – B)

A = 90° – B ( Both A and B are acute angles)

⇒ A + B = 90°.

11. If sec A = 2x and tan A = 2 , find the value of 2  x2 − 1  .
x  x2 

Sol. 2  x 2 − 1  = 2 sec2 A tan2 A = 2 (sec2 A − tan2 A) = 1 × 1 = 1
 x2   4 − 4  4 2 2



12. In a DABC, if ∠C = 90°, prove that sin2 A + sin2 B = 1.

Sol.  ∠C = 90°

\ ∠A + ∠B = 180° – ∠C = 90°
Now, sin2 A + sin2 B = sin2 A + sin2 (90° – A) = sin2 A + cos2 A = 1

13. If sec 4A = cosec (A – 20°) where 4 A is an acute angle, find the value of A. [NCERT]

Sol. We have sec 4 A = cosec (A – 20°)

⇒ cosec (90° – 4 A) = cosec (A – 20°)

\ 90° – 4 A = A – 20°

⇒ 90° + 20° = A + 4 A ⇒ 110° = 5 A

\ A = 110 = 22°.
5
14. Evaluate sin 25° cos 65° + cos 25° sin 65°. [NCERT]

Sol. sin 25° . cos 65° + cos 25° . sin 65°

= sin (90° – 65°) . cos 65° + cos (90° – 65°) . sin 65°

= cos 65° . cos 65° + sin 65° . sin 65°
= cos2 65° + sin2 65° = 1.

Short Answer Questions-II [3 marks]

1. If sin A = 3 , calculate cos A and tan A. [NCERT]
4
Sol. Let us first draw a right DABC in which ∠C = 90°.

Now, we know that

sin A = PHeryppeontdenicuusl=ear BA=CB 3
4

Let BC = 3k and AB = 4k, where k is a positive number.

Then, by Pythagoras Theorem, we have

AB2= BC2 + AC2 ⇒ (4k)2 = (3k)2 + AC2 Fig. 10.3

⇒ 16k2 – 9k2 = AC2 ⇒ 7k2 = AC2

\ AC = 7k
\ cos A =
AC = 7k = 7 and tan=A BA=CC 3k = 3
AB 4k 4 7k 7

2. Given 15 cot A = 8, find sin A and sec A. [NCERT]
Sol. Let us first draw a right DABC in which ∠B = 90°.
Now, we have, 15 cot A = 8

252 Xam idea Mathematics–X

\ cot A = =185 BA=CB Base
Perpendicular

Let AB = 8k and BC = 15k

Then, AC = ( AB)2 + (BC)2 (By Pythagoras Theorem)

= (8k)2 + (15k)2 = 64k2 + 225k2 = 289k2 = 17k

PHeryppeontdenicuus=lear BA=CC 15k 15 Fig. 10.4
17k 17
\ sin A = =

and, sec A = Hypotenuse AA=CB 17k = 17
Base = 8k 8

3. In Fig. 10.5, find tan P – cot R. [NCERT]

Sol. Using Pythagoras Theorem, we have

PR2 = PQ2 + QR2

⇒ (13)2 = (12)2 + QR2

⇒ 169 = 144 + QR2

⇒ QR2 = 169 – 144 = 25 ⇒ QR = 5 cm

QR 1=52 and cot R QR = 5
Now, ta=n P = PQ PQ 12

\ tan P – cot R = 5 − 5 = 0
12 12

4. If sin q + cos q = 3 , then prove that tan q + cot q = 1. [NCERT Exemplar]

Sol. sin q + cos q = 3

⇒ (sin q + cos q)2 = 3 ( sin2 q + cos2 q = 1)
⇒ sin2 q + cos2 q + 2 sin q cos q = 3

⇒ 2 sin q cos q = 2
⇒ sin q . cos q = 1 = sin2 q + cos2 q

⇒ 1= sin2 θ + cos2 θ
sin θ cos θ

⇒ 1 = tan q + cot q
Therefore tan q + cot q = 1.

5. Prove that : 1 − sin θ = (sec q – tan q)2
1 + sin θ

Sol. LHS = 1 − sin θ
1 + sin θ

= 1 − sin θ × 1 − sin θ (Rationalising the denominator)
1 + sin θ 1 − sin θ

= (1 – sin i)2 = d 1 – sin i 2 = d 1 – sin i 2
1 – sin2 i cos i cos cos i
n i n

= (sec q – tan q)2 = RHS

Introduction to Trigonometry 253

6. Without using tables, evaluate

cosesce2c25547° °−−cotatn2 36° + 2sin2 38° . sec2 52° – sin2 45°.
2 33°

Sol. We have, sec2 54° − cot2 36° + 2 sin2 38° . sec2 52° – sin2 45°
cos ec2 57° − tan2 33°

= sec2 (90° − 36°) − cot2 36° + 2 sin2 38° . sec2 (90° – 38°) – sin2 45°
cos ec2 (90° − 33°) − tan2 33°

= cos ec2 36° − cot2 36° + 2 sin2 38° . cosec2 38° –  1 2
sec2 33° − tan2 33°  
 2 

= 1 + 2.1 − 1 = 3 − 1 = 5
1 2 2 2

7. Prove that : (1 + cot A – cosec A)(1 + tan A + sec A) = 2 [CBSE 2019 (30/1/2)]

Sol. LHS  = (1 + cotA – cosecA) (1 + tanA +secA)

= c1 + cos A – 1 A mc1 + sin A + 1 m
sin A sin cos A cos A

= d sin A + cos A–1 nd cos A + sin A + 1 n
sin A cos A

= sin 1 A 6(sin A+ cos A – 1)@(sin A+ cos A +1)
A cos

= sin 1 A 7(sin A + cos A) 2 – 1A
A cos

= sin 1 A 6sin2 A+ cos2 A+ 2 sin A+ cos A – 1@
A cos

= sin 1 A ^1 + 2 sin A cos A – 1h
A cos

= 2 sin A cos A =2= RHS.
sin A cos A

8. If sin 3q = cos (q – 6°) where 3q and (q – 6°) are both acute angles, find the value of q.

Sol. According to question:

sin 3q = cos (q – 6°)

⇒ cos (90° – 3q) = cos (q – 6°) ( cos (90° – q) = sin q)

⇒ 90° – 3q = q – 6° (comparing the angles)

⇒ 4q = 90° + 6° = 96° ⇒ q= 96 = 24°
4
Hence, q = 24°
1
9. If sec q = x + 1 , prove that sec q + tan q = 2x or 2x .
4x

Sol. Let sec q + tan q = l ...(i)

We know that, sec2 q – tan2 q = 1

(sec q + tan q) (sec q – tan q) = 1 ⇒ l(sec q – tan q) = 1
1
sec q – tan q = λ ...(ii)

Adding equations (i) and (ii), we get

254 Xam idea Mathematics–X

2 sec q = l + 1 ⇒ 2  x + 1  = λ + 1
λ  4x  λ

⇒ 2x + 1 = l+ 1
2x λ
1
On comparing, we get l = 2x or l = 2x

⇒ sec q + tan q = 2x or 1
2x
Alternative Method

We have sec q= x + 1
4x

tan=2 q sec2 q −1

= x2 + 1 + 1 −1
16x2 2

=x2 + 1 − 1
16x2 2

=  x − 1 2
 4x 

⇒ tan q = ±  x − 1 
 4x 

sec + tan θ is given by

x + 1 + x− 1 or x + 1 − x+ 1
4x 4x 4x 4x
1
= 2x or 2x

10. Find an acute angle q, when cos θ − sin θ = 1− 3 .
cos θ + sin θ 1+ 3

Sol. We have, cos θ − sin θ

cos θ − sin θ = 1− 3 ⇒ cos θ = 1 − 3
cos θ + sin θ 1+ 3 cos θ + sin θ 1 + 3

cos θ

(Dividing numerator & denominator of the LHS by cos q)

⇒ 1− tan θ = 1− 3
1+ tan θ 1+ 3
On comparing we get

⇒ tan q = 3 ⇒ tan q = tan 60° ⇒ q = 60°

11. The altitude AD of a DABC, in which ∠A is an obtuse angle has length 10 cm. If BD = 10 cm

and CD = 10 3 cm, determine ∠A.

Sol. DABD is a right triangle right angled at D, such that AD = 10 cm and BD = 10 cm.

Let ∠BAD = q BD ⇒ tan q = 10 =1
\ tan q = AD 10

⇒ tan q = tan 45° ⇒ q = ∠BAD = 45° ...(i)

DACD is a right triangle right angled at D such that AD = 10 cm and DC = 10 3 cm.

Introduction to Trigonometry 255

Let ∠CAD = φ

\ tan φ = CD ⇒ tan φ = 10 3 = 3
AD 10

⇒ tan φ = tan 60° ⇒ φ = ∠CAD = 60° ...(ii)

From (i) & (ii), we have

∠BAC = ∠BAD + ∠CAD = 45° + 60° = 105°

12. If cosec q = 13 , evaluate 2 sin θ − 3 cos θ . Fig. 10.6
12 4 sin θ − 9 cos θ

Sol. Given cosec q = 13 , then sin q = 12
12 13

cos2 q = 1 – sin2 q = 1 – 12 2 = 169 −144 = 25 = 5
 13  169 169 13

2 sin i – 3 cos i 2 # 12 – 3 # 5 24 – 15 9
4 sin i – 9 cos i 13 13 48 – 45 3
Now, = = = = 3
12 5
4 # 13 – 9 # 13

13. Prove that : tan i – cot i = cos i + sin i [CBSE 2019 (30/5/1)]
1 – tan i 1 – cot i cos i – sin i

Sol. LHS = tan i – 1 cot i i
1 – tan i – cot

sin i cos i

= cos i – sin i = sin i + cos i
cos i – sin i cos i – sin i
1– sin i 1– cos i
cos i sin i

= sin i + cos i = cos i + sin i = RHS
cos i – sin i cos i – sin i

14. Prove that : cos3 θ+ sin3 θ + cos3 θ − sin3 θ = 2
cos θ+ sin θ cos θ − sin θ
Sol. We have,

LHS = cos3 θ + sin3 θ + cos3 θ − sin3 θ
cos θ + sin θ cos θ − sin θ

= (cos θ + sin θ) (cos2 θ + sin2 θ − cos θ . sin θ) + (cos θ − sin θ) (cos2 θ + sin2 θ + cos θ . sin θ)
cos θ + sin θ cos θ − sin θ

= cos2 q + sin2 q – cos q . sin q + cos2 q + sin2 q + cos q . sin q

= 1 – cos q . sin q + 1 + cos q . sin q ( sin2 q + cos2 q = 1)

= 2 = RHS 2 cos2 θ − 1
sin θ cos θ
15. Prove that: cot q – tan q =

Sol. LHS = cot q – tan q = cos θ − sin θ
sin θ cos θ

= cos2 T sin2 T cos2 T (1 cos2 T)
sin T cos T sin T cos T

256 Xam idea Mathematics–X

= cos2 θ − 1 + cos2 θ = 2cos2 θ − 1 = RHS
sin θ cos θ sin θ cos θ

16. Prove that: 1 + cot 2 α α = cosec a [NCERT Exemplar]
1+ cosec

Sol. We have, cot2 α
1 + cosec
LHS = 1 + α

=1+ cot2 α(1 − cosec α) α) (Multiply numerator and denominator by (1 – cosec a))
(1 + cosec α)(1 − cosec

=1+ cot2 a(1 – cosec a)
1 – cosec2 a

=1+ cot2 α(1 − cosec α) ( cosec2 a = 1 + cot2 a)
1 − (1 + cot2 α)

=1+ cot2 α(1 − cos ec α) =1+ cot2 α(1 − cos ec α)
1 − 1 − cot2 α − cot2 α

= 1 – (1 – cosec a) = 1 – 1 + cosec a = cosec a = RHS

17. If cos i + sin i = 2 cos i, show that cos i – sin i = 2 sin i . [CBSE 2019 (30/5/1)]

Sol. Given, cos i + sin i = 2 cos i

Now, (cos i + sin i)2 + ]cos i – sin ig2

cos2 i + sin2 i + 2 cos i. sin i + cos2 i + sin2 i – 2 cos i. sin i

1+1=2

Again, (cos i + sin i)2 + ]cos i – sin ig2 = 2

& ^ 2 cos ih2 + ]cos i – sin ig2 = 2

& 2 cos2 i + (cos i – sin i)2 = 2 ⇒ (cos i – sin i)2 = 2 – 2 cos2 i = 2 (1 – cos2 i) = 2 sin2 i

` cos i – sin i = 2 sin i Hence proved.

18. If tan (A +B) = 3 and tan (A – B) = 1 ; 0° < A + B ≤ 90°; A > B, find A and B. [NCERT]
3

Sol. We have, tan (A + B) = 3 ⇒ tan (A + B) = tan 60°

\ AtAan+– B(BA=–=B360)0°= ° 13 ⇒ tan (A – B ) = tan 30° ……((iii))
Again,
\

Adding (i) and (ii), we have

2A = 90° ⇒ A = 45°

Putting the value of A in (i), we have

45° + B = 60°

\ B = 60° – 45° = 15°

Hence, A = 45° and B = 15°. of a DABC, then show that sin  B+C  = cos A . [NCERT]
19. If A, B and C are interior angles  2  2

Sol. Since A, B and C are the interior angles of a DABC,

Therefore, A + B + C = 180° ⇒ A + B + C = 180°
2 2

Introduction to Trigonometry 257

⇒ A + (B + C) = 90° ⇒ B+ C = 90° − A
2 2 2 2

Now, taking sin on both sides, we have

sin  B + C = sin  90° − A  ⇒ sin B+ C = cos A
 2   2  2 2

20. Prove that: (cosec q – cot q)2 = 1 − cos θ
1+ cos θ

Sol. LHS = (cosec q – cot q)2

= 1 − cos θ 2 = 1 − cos θ 2
 sin θ sin θ   sin θ 

= (1 − cos θ)2 = (1 − cos θ)2
sin2 θ 1 − cos2 θ

= (1 – (1 – cos i)2 = 1– cos i = RHS
cos i) (1+ cos i) 1+ cos i

21. Prove that: sin θ − 2 sin3 θ = tan θ [NCERT]
2 cos3 θ − cos θ

sin T 2 sin3 T sin T (1 2 sin2 T) sin T ª 1 2sin2 T º
Sol. LHS = 2 cos3 T cos T « 1¼»
cos T (2 cos2 T 1) cos T ¬ 2(1 sin2 T)

= tan θ  1 − 2 sin2 θ  = tan θ  1 − 2 sin2 θ = tan θ = RHS.
 − 2 sin2 θ −   1 − 2 sin2 
 2 1   θ 

22. Prove that: (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

[NCERT, CBSE 2019 (30/1/2)]
Sol. LHS = (sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + cosec2 A + 2sin A . cosec A + cos2 A + sec2 A + 2 cos A. sec A

= (sin2 A + cosec2 A + 2) + (cos2 A + sec2 A + 2) §sin A . cosec A 1·
= (sin2 A + cos2 A) + (cosec2 A + sec2 A) + 4 ¨ ¸
= 1 + 1 + cot2 A + 1 + tan2 A + 4 © cos A . sec A 1 ¹

= 7 + tan2 A + cot2 A = RHS ( 1 + cot2 A = cosec2 A and 1 + tan2 A = sec2 A)

23. Prove that: 1 + cos θ − sin2 θ = cot θ [NCERT]
sin θ (1+ cos θ)

Sol. LHS = 1 + cos θ − sin2 θ
sin θ (1 + cos θ)

To obtain cot q in RHS, we have to convert the numerator of LHS in cosine function and
denominator in sine function.

Therefore converting sin2 q = 1 – cos2 q, we get

= 1 + cos θ − (1 − cos2 θ) = 1 + cos θ −1 + cos2 θ = cos θ + cos2 θ
sin θ (1 + cos θ) sin θ(1 + cos θ) sin θ (1 + cos θ)

= scions θ (cos θ + 1) = cos θ = cot θ = RHS
θ (1 + cos θ) sin θ

258 Xam idea Mathematics–X

24. Prove that: 1 A−1+ cos 1 A+1 = cosec A + sec A
cos A + sin A + sin

Sol. LHS = cos 1 A−1 + cos 1 A+1
A + sin A + sin

cos A + sin A + 1 + cos A + sin A−1
= (cos A + sin A − 1)(cos A + sin A + 1)

2(cos A + sin A) 1 = cos2 2(cos A + sin A) A−1
= (cos A + sin A)2 − A + sin2 A + 2cos A sin

= ccoossAA+sisninAA = cos A A + sin A A = 1 A + 1 A
cos A sin cos A sin sin cos

= cosec A + sec A = RHS.

25. Prove that: cos A + 1 + sin A = 2 sec A [NCERT]
+ sin cos A
1 A

Sol. LHS = 1 cos A + 1 + sin A
+ sin A cos A

= co(s12 +Asi+n(1A+) csoins A)2 = cos2 A + 1 + sin2 A + 2 sin A
A (1 + sin A) cos A

= (cos2 A + sin2 A) + 1 + 2 sin A = 1+1+ 2 sin A
(1 + sin A) cos A (1 + sin A) cos A

= (1 +2(1sin+ sin A) A= 2 = 2 sec A = RHS.
A) cos cos A

26. Without using trigonometric tables, prove that:

sec2 i – cot2 (90° – i) + (sin2 40° + sin2 50°) = 2
cosec2 67° – tan2 23°

Sol. We have, sec2 i – cot2 (90° – i)
cosec2 67° – tan2 23°
LHS = + (sin2 40° + sin2 50°)

= cos ec2 sec2 θ − tan2 θ 23° + {sin2 40° + sin2 (90° − 40°)}
(90° − 23°) − tan2

= secs2ec223θ° − tan2 θ + (sin2 40° + cos2 40°) = 1 +1 = 2 = RHS
− tan2 23° 1

27. Evaluate: cosec2 (90° – i) – tan2 i – 2 tan2 30° sec2 37° sin2 53° [CBSE 2019 (30/5/2)]
2 (cos2 37° + cos2 53°) cosec2 63° – tan2 27°

Sol. We have, cosec2 ]90 – ig – tan2 i – 2 tan2 30° sec2 37° sin2 53°
2 (cos2 37° + cos2 53°) cosec2 63° – tan2 27°

sec2 i – tan2 i 2×e 1 2 × sec2 (90° – 53°) . sin2 53°
(90° – 53°) + cos2 3
o

= 2 (cos2 53°) – cosec2 (90° – 27°) – tan2 27°

= 2 1 – 2× 1 × cosec2 53° × sin2 53°
53° + 3

(sin2 cos2 53°) sec2 27° – tan2 27°

Introduction to Trigonometry 259

= 2×1 1 – 2× 1 ×1 = 1 – 2 = 3 –4 = – 1
3 2 3 6 6

1

Long Answer Questions [5 marks]

1. In DPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of

sin P, cos P and tan P. [NCERT]

Sol. We have a right-angled DPQR in which ∠Q = 90°.

Let QR = x cm

Therefore, PR = (25 – x) cm

By Pythagoras Theorem, we have

PR2 = PQ2 + QR2

⇒ (25 – x)2 = 52 + x2 ⇒ (25 – x)2 – x2 = 25

⇒ (25 – x – x) (25 – x + x) = 25 Fig. 10.7
⇒
(25 – 2x) 25 = 25 ⇒ 25 – 2x = 1

⇒ 25 – 1 = 2x ⇒ 24 = 2x

\ x = 12 cm

Hence, QR = 12 cm

PR = (25 – x) cm = 25 – 12 = 13 cm

PQ = 5 cm

\ sin P = QR = 12 ; cos P = PQ = 5 ; tan P = QR = 12
PR 13 PR 13 PQ 5

2. In triangle ABC right-angled at B, if tan A = 1 find the value of:
3

(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C. [NCERT]

Sol. We have a right-angled DABC in which ∠B = 90° and tan A = 1
3
1 BC
Now, tan A = 3 = AB

Let BC = k and AB = 3k
\ By Pythagoras Theorem, we have

⇒ AC2 = AB2 + BC2

⇒ AC2 = ( 3k)2 + (k)2 = 3k2 + k2 Fig. 10.8
⇒
AC2 = 4k2 ⇒ AC = 2k

Now, sin A = Perpendicular 2=kk 1 ; cos A = HypBoatesenu=se =23kk 3
Hypotenuse= 2 2

sin C = Perpendicular =23kk 3 ; cos C = HypBoatesenus=e 2=kk 1
Hypotenu=se 2 2

(i) sin A . cos C + cos A . sin C = 1 × 1 + 3 × 3 = 1 + 3 = 4 =1
2 2 2 2 4 4 4

(ii) cos A . cos C – sin A. sin C = 3 × 1 − 1 × 3 = 3 − 3 = 0.
2 2 2 2 4 4
260 Xam idea Mathematics–X

3. If cot q = 7 , evaluate: (i) (1 + sin θ ) (1 − sin θ) (ii) cot2 q.
8 (1 + cos θ ) (1 − cos θ)

Sol. Let us draw a right triangle ABC in which ∠B = 90° and ∠C = q.

We h=ave, cot q = 78 P=erpeBnadseicular BC (given)
AB

Let BC = 7k and AB = 8k
Therefore, by Pythagoras Theorem
AC2 = AB2 + BC2 = (8k)2 + (7k)2 = 64k2 + 49k2

AC2 = 113k2 \ AC = 113k Fig. 10.9

\ sin q = Perpendicular AA=CB 8k = 8
Hypotenus=e 113k 113

and cos q = HypBoatesenu=se BA=CC 7k = 7
113k 113

(i) ((11++ sin θ) (1 − sin θ) = 1− sin2 θ = 1 −  8 2
cos θ) (1 − cos θ) 1− cos2 θ 1 −  113 
 7 2
 113 

= 1 − 64 = 113 − 64 = 49
Alternate method: 1 − 113 113 64
49
113 113 − 49
113

(1 + sin θ)(1 − sin θ) = 1 − sin2 θ = cos2 θ = cot2 q =  7 2 = 49
(1 + cos θ)(1 − cos θ) 1 − cos2 θ sin2 θ  8  64

(ii) cot2 q =  7 2 = 49 .
 8  64
A
4. Prove that : tan2 – 1 + cosec2 A = 1 – 1 [CBSE 2019 (30/5/1)]
tan2 A sec2 A – cosec2 A 2 cos2 A
tan2 A cosec2 A
Sol. LHS = tan2 A – 1 + sec2 A – cosec2 A

sin2 A 1

= cos2 A + sin2 A

sin2 A –1 1 – 1
cos2 A cos2 A sin2 A

sin2 A 1

= cos2 A + sin2 A
sin2 A – cos2 A sin2 A – cos2 A

cos2 A sin2 A. cos2 A

= sin2 A + cos2 A = sin2 A+ cos2 A = sin2 A 1
sin2 A – cos2 A sin2 A – cos2 A sin2 A – cos2 A – cos2 A

= 1 – 1 – cos2 A = 1 – 1 = RHS
cos2 A 2 cos2 A

Introduction to Trigonometry 261

5. Write all the other trigonometric ratios of ∠A in terms of sec A. [NCERT]
Fig. 10.10
Sol. Let us consider a right-angled DABC in which ∠B = 90°.
For ∠A we have [NCERT]
Base = AB, Perpendicular = BC and Hypotenuse = AC

\ sec A = Hypotenuse = AC
Base AB

⇒ sec A = AC ⇒ AC = AB sec A
1 AB

Let AB = k and AC = k sec A

\ By Pythagoras Theorem, we have

AC2 = AB2 + BC2 ⇒ k2 sec2 A = k2 + BC2

\ BC2 = k2 sec2 A – k2 ⇒ BC = k sec2 A −1

\ sin A = BACC = k sec2 A −1 = sec2 A −1
k sec A sec A

cos A == AACB k=sekc A 1
sec A

tan A = BACB = k sec2 A−1 = sec2 A −1
k

cot 1 A = 1
A = tan sec2 A −1

cosec A = AC = k k sec A sec A
BC = sec2 A −1

sec2 A −1

6. Prove that: 1 + tan2 A  =  1 − tan A 2 = tan2 A
 cot 2 A   1 − cot 
 1 +   A 

Sol. LHS =  1 + tan2 A = sec2 A
1 + cot2 A  cosec2 A

1 cs=ions22 AA tan2 A
= c=os12 A

sin2 A

RHS =  1 – tan A 2 =  tan A 2
1 – cot A  1– 1 A 
 
1 – tan 

§ ·2 1 tan A ·2
¨ 1 tan A ¸ tan A 1 ¹¸
= ¨ tan A 1 ¸ § u tan A = (– tan A)2 = tan2 A
¸¹¸ ©¨
¨¨© tan A

LHS = RHS.

7. Prove that: tan2 A – tan2 B = cos2 B − cos2 A = sin2 A − sin2 B
cos2 B cos2 A cos2 A cos2 B

Sol. LHS = tan2 A – tan2 B = sin2 A sin2 B
cos2 A − cos2 B

262 Xam idea Mathematics–X

= sin2 A cos2 B – cos2 A sin2 B = (1 – cos2 A) cos2 B – cos2 A (1 – cos2 B)
cos2 A cos2 B cos2 A cos2 B

= cos2 B − cos2 Acos2 B − cos2 A + cos2 Acos2 B = cos2 B − cos2 A
cos2 Acos2 B cos2 Acos2 B

Also cos2 B − cos2 A = (1 − sin2 B) − (1 − sin2 A)
cos2 Acos2 B cos2 Acos2 B

= sin2 A − sin2 B = RHS.
cos2 Acos2 B

8. Prove that: cosec A cosec A = 2 + 2 tan2 A = 2 sec2 A
cosec A − 1 + cosec A + 1

Sol. LHS = cosec A + cosec A 1)
(cosec A − 1) (cosec A +

= cosec A (cosec A + 1) + cosec A (cosec A − 1)
(cosec A − 1) (cosec A + 1)

= cosec2 A+ cosec A+ cosec2 A– cosec A = 2 cosec2 A = 2 cosec2 A
(cosec2 A – 1) 1+ cot2 A –1 cot2 A

= 2 cosec2 A tan2 A = 2(1 + cot2 A) . tan2 A

= 2 tan2 A + 2 tan2 A . cot2 A ( tan A. cot A = 1)

= 2 + 2 tan2 A = 2(1 + tan2 A) = 2 sec2 A = RHS.

9. Prove that: (sin q + sec q)2 + (cos q + cosec q)2 = (1 + sec q cosec q)2

Sol. LHS = (sin q + sec q)2 + (cos q + cosec q)2

=  sin θ + 1 2 +  cos θ + 1 2 =  sin θ cos θ + 1 2 +  cos θ sin θ + 1 2
 cos   sin   cos θ   sin θ 
 θ   θ     

= (sin θ cos θ + 1)2 + (cos θ sin θ + 1)2 = (sin θ cos θ + 1)2 1 + 1 
cos2 θ sin2 θ  sin2 
 cos2 θ θ 

= (sin θ cos θ + 1)2  sin2 θ + cos2 θ  = (sin θ cos θ + 1)2 .  1 
   θ sin2 
 cos2 θ sin2 θ   cos2 θ 

=  sin θ cos θ + 1 2 = 1 + 1 2
   θ sin 
 cos θ sin θ  cos θ 

= (1 + sec q cosec q)2 = RHS.

10. Prove that: ]cosec 1 cot xg – 1 x = 1 x – 1
x+ sin sin (cosec x – cot x)

Sol. In order to show that,

1 – 1 x = 1 – (cosec 1 cot x)
]cosec x + cot xg sin sin x x–

It is sufficient to show

cosec 1 cot x + cosec 1 cot x = 1 x + 1 x
x+ x– sin sin

⇒ (cosec 1 cot x) + (cosec 1 cot x) = 2 x ...(i)
x+ x− sin

Introduction to Trigonometry 263

Now, LHS of above is

(cosec 1 cot x) + (cosec 1 cot x)
x+ x−

= (cosec x − cot x) + (cosec x + cot x)
(cosec x − cot x) (cosec x + cot x)

= 2 cosec x x ( (a + b) (a – b) = a2 – b2)
cosec 2 x − cot2

= 2 cosec x = 2 = RHS of (i)
1 sin
x

Hence, 1 + 1 cot x) = 1 + 1
]cosec x + cot xg (cosec x – sin x sin x

or (cosec 1 cot x) − 1 x = 1 x − (cosec 1 cot x) .
x+ sin sin x−

11. Prove that: cosec T cot T = (cosec q + cot q)2 = 1 + 2 cot2 q + 2 cosec q cot q
cosec T cot T

Sol. LHS = cosec θ + cot θ
cosec θ − cot θ

Rationalising the denominator, we get

= (cosec θ + cot θ) × (cosec θ + cot θ)
(cosec θ − cot θ) (cosec θ + cot θ)

= (cosec θ + cot θ)2 = (cosec θ + cot θ)2 ( cosec2 q – cot2 q = 1)
cosec 2 θ − cot2 θ 1

= cosec2 q + cot2 q + 2 cosec q . cot q

= (1 + cot2 q) + cot2 q + 2 cosec q . cot q

= 1 + 2 cot2 q + 2 cosec q . cot q = RHS.

12. Prove that: 2 sec2 q – sec4 q – 2 cosec2 q + cosec4 q = cot4 q – tan4 q

Sol. LHS = 2 sec2 q – sec4 q – 2 cosec2 q + cosec4 q

= 2 (sec2 q) – (sec2 q)2 – 2 (cosec2 q) + (cosec2 q)2

= 2 (1 + tan2 q) – (1 + tan2 q)2 – 2(1 + cot2 q) + (1 + cot2 q)2

= 2 + 2 tan2 q – (1 + 2 tan2 q + tan4 q) – 2 – 2 cot2 q + (1 + 2 cot2 q + cot4 q)

= 2 + 2 tan2 q – 1 – 2 tan2 q – tan4 q – 2 – 2 cot2 q + 1 + 2 cot2 q + cot4 q

= cot4 q – tan4 q = RHS

13. Prove that : ]1 + cot i + tan ig]sin i – cos ig = sin2 i cos2 i [CBSE 2019 (30/5/1)]
^sec3 i – cosec3 ih

Sol. LHS = ]1+ cot i + tan ig]sin i – cos ig
(sec3 i – cosec3 i)

d1+ cos i + sin i n]sin i – cos ig
sin i cos i
=
1 1
d cos3 i – sin3 i n

= ^sin i. cos i + cos2 i + sin2 ih]sin i – cos ig
sin i. cos i

sin3 i – cos3 i

sin3 i. cos3 i

264 Xam idea Mathematics–X

= sin3 i – cos3 i × sin3 i . cos3 i
sin i. cos i sin3 i – cos3 i

= sin2 i. cos2 i = RHS

HOTS [Higher Order Thinking Skills]

1. Prove that: 1 tan i i + 1 cot i i = 1 + sec q cosec q = 1 + tan q + cot q
– cot – tan
sin i cos i

Sol. LHS = tan i + cot i = cos i + sin i
– cot i – tan
1 1 i 1 – cos i 1 – sin i
sin i cos i

= sin i # sin i + cos i # cos i
cos i (sin i – cos i) sin i (cos i – sin i)

= cos T sin2 T cos T) cos2 T
(sin T sin T { (sin T cos T)}

= cos θ sin2 θ cos θ) − sin θ cos2 θ cos θ) = sin3 θ − cos3 θ
(sin θ − (sin θ − cos θ (sin θ − cos θ) sin θ

= (sin θ − cos θ) (sin2 θ + cos2 θ + sin θ cos θ) = 1 + sin θ cos θ
cos θ sin θ (sin θ − cos θ) sin θ cos θ

= 1 + sin θ cos θ = 1 . 1 +1 ...(i)
sin θ cos θ sin θ cos θ sin θ cos θ

= sec q cosec q + 1 = RHS ...(ii)

For second part:

Now from (i), we have

LHS = 1 +1
sin θ cos θ

Putting 1 = sin2 q + cos2 q

= sin2 θ + cos2 θ +1= sin2 θ + cos2 θ +1
sin θ cos θ sin θ cos θ cos θ sin θ

= sin θ + cos θ + 1 = tan q + cot q + 1 = RHS
cos θ sin θ

2. If tan A = n tan B and sin A = m sin B, prove that cos2 A = m2 –1 .
n2 –1
Sol. We have to find cos2 A in terms of m and n. This means that the angle B is to be eliminated from

the given relations.

Now, tan A = n tan B

⇒ tan B = 1 tan A ⇒ cot B = n A
n tan

and sin A = m sin B

⇒ sin B = 1 sin A ⇒ cosec B = m
m sin A

Substituting the values of cot B and cosec B in cosec2 B – cot2 B = 1, we get

⇒ m2 A − n2 A =1 ⇒ m2 A − n2 cos2 A =1
sin2 tan2 sin2 sin2 A

Introduction to Trigonometry 265

⇒ m2 − n2 cos2 A =1 ⇒ m2 – n2 cos2 A = sin2 A
sin2 A

⇒ m2 – n2 cos2 A = 1 – cos2 A

⇒ m2 – 1 = n2 cos2 A – cos2 A m2 −1 = cos2 A
⇒ m2 – 1 = (n2 – 1) cos2 A ⇒ n2 −1

3. Prove the following identity, where the angle involved is acute angle for which the expressions
are defined.

cos A − sin A+1 = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A.
cos A + sin A−1

cos A − sin A + 1

Sol. LHS = cos A − sin A+1 = cos sin A A −1 = cot A−1+ cosec A
cos A + sin A−1 A + sin cot A+1− cosec A

sin A

= (cot A+ cosec A) − (cosec 2 A − cot2 A) [ cosec2 A – cot2 A = 1]
cot A − cosec A + 1

= (cot A + cosec A) − [(cosec A + cot A) (cosec A − cot A)]
cot A − cosec A + 1

(cosec A + cot A) (1 − cosec A + cot A)
= (cot A − cosec A + 1)

= cosec A + cot A = RHS.

4. If x sin3 q + y cos3 q = sin q cos q and x sin q = y cos q, prove x2 + y2 = 1.

Sol. We have, x sin3 q + y cos3 q = sin q cos q
⇒ (x sin q) sin2 q + (y cos q) cos2 q = sin q cos q

⇒ x sin q (sin2 q) + (x sin q) cos2 q = sin q cos q ( x sin q = y cos q)

⇒ x sin q (sin2 q + cos2 q) = sin q cos q

⇒ x sin q = sin q cos q ⇒ x = cos q

Now, we have x sin q = y cos q

⇒ cos q sin q = y cos q ( x = cos q)

⇒ y = sin q

Hence, x2 + y2 = cos2 q + sin2 q = 1.

5. If tan q + sin q = m and tan q – sin q = n, show that (m2 – n2) = 4 mn .

Sol. We have, given tan q + sin q = m, and tan q – sin q = n, then

LHS = (m2 – n2) = (tan q + sin q)2 – (tan q – sin q)2

= tan2 q + sin2 q + 2 tan q sin q – tan2 q – sin2 q + 2 tan q sin q

= 4 tan q sin q = 4 tan2 i sin2 i

= 4 sin2 θ (1 − cos2 θ) = 4 sin2 θ − sin2 θ
cos2 θ cos2 θ

= 4 tan2 θ − sin2 θ = 4 (tan θ − sin θ)(tan θ + sin θ) = 4 mn = RHS

266 Xam idea Mathematics–X

6. If cosec q – sin q = l and sec q – cos q = m, prove that l2 m2 (l2 + m2 + 3) = 1.

Sol. LHS = l2 m2 (l2 + m2 + 3)

= (cosec q – sin q)2 (sec q – cos q)2 {(cosec q – sin q)2 + (sec q – cos q)2 + 3}

=  1 θ − sin θ 2  1 − cos θ 2  1 − sin θ 2 +  1 − cos 2 + 3
 sin   cos θ   sin θ   cos θ θ 
     


= 1 − sin2 θ 2 1 − cos2 θ 2  1 − sin2 θ 2 + 1 − cos2 θ 2 + 3
 sin θ   cos θ   sin θ   cos θ  
      

=  cos2 θ 2  sin2 θ 2  cos2 θ 2 +  sin2 θ 2 + 3
 sin θ   cos θ   sin θ   cos θ  
       

= cos4 θ sin4 θ cos4 θ + sin4 θ + 3 = cos2 q sin2 q  cos6 θ + sin6 θ + 3cos2 θsin2 θ 
sin2 θ  cos2 θ   cos2 θ sin2 θ 
cos2 θ  sin2 θ  

= cos6 q + sin6 q + 3 cos2 q sin2 q = [(cos2 q)3 + (sin2 q)3] + 3 cos2 q sin2 q

= [(cos2 q + sin2 q)3 – 3 cos2 q sin2 q (cos2 q + sin2 q)] + 3 cos2 q sin2 q

( a3 + b3 = (a + b)3 – 3ab (a + b))

= 1 – 3 cos2 q sin2 q + 3 cos2 q sin2 q ( cos2 q + sin2 q = 1)

= 1 = RHS.

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]
1. Choose and write the correct option in each of the following questions.

(i) Value of sin 30° tan 45° is
3
(a) 2 (b) 2 (c) 1 (d) 0
2

(ii) If sin i + cos i = 2 cos i, (i ! 90°), then the value of tan q is

(a) 2 –1 (b) 2 + 1 (c) 2 (d) – 2

(iii) If 2x = secq and 2 = tanq, then the value =x2 – 1 is
x x2 G
1 1
(a) 4 (b) 4 (c) 2 (d) 2

(iv) If sin 77° = x, then the value of tan 77° is
1 x x
(a) 1+ x2 (b) 1 + x2 (c) 1– x2 (d) None of these

(v) The value of 4 (sin430° + cos4 60°) – 3 (cos2 45° – sin2 90°) is
1
(a) 2 (b) 1 (c) 2 (d) 3

2. Fill in the blanks.

(i) The value of cos 54° – sin 36° is _____________ .

(ii) The value of sin q cos (90° – q) + sin(90° – q) cosq is _____________ .

(iii) If tan A= 4 , then cot A = _____________ .
3

Introduction to Trigonometry 267

(iv) The value of sec 30° is _____________ .
cosec 60°

(v) If 2 cos3q = 1 then, q = _____________ . [3q < 90°]

QQ Very Short Answer Questions : [1 mark each]
[CBSE 2018 (30/1)]
3. What is the value of (cos2 67° – sin2 23°)? [CBSE 2019 (30/1/1)]
[CBSE 2019 (30/1/1)]
4. Find A if tan 2A = cost (A – 24°)

5. Find the value of (sin2 33° + sin2 57°)

6. Evaluate: [CBSE 2019 (30/2/1)]
sin2 60° + 2 tan 45° – cos2 30°


7. If sin A= 3 , calculate sec A. [CBSE 2019 (30/2/1)]
4 [CBSE 2019 (30/3/1)]
5
8. If tan a = 12 , find the value of sec a.

9. Evaluate : [CBSE 2019 (30/4/2)]
tan 65°
cot 25°

10. Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
[CBSE 2019 (30/4/2)]

11. Evaluate : tan 36° [CBSE 2019(C)(30/1/1)]
cot 54° [CBSE 2019(C)(30/1/1)]

12. If cosec2q(1 + cosq)(1 – cosq) = k, then find the value of k

13. If sin q = 12 , then find cosec q.
13
14. Find the value of (sin2 48° + cos2 48°).

15. If tan a = 3 and tan b = 1 , then find the value of cot(a + b).
3
16. If DABC is right angled at C, then find the value of cos (A +B). [NCERT Exemplar]

QQ Short Answer Questions-I: [2 marks each]

17. A, B, C are interior angles of DABC. Prove that cosec d A+B n= sec C . [CBSE 2018 (C)(30/1)]
2 2

18. If tan 3A = cos (A – 26°), 3A < 90°, then find the value of ∠A.

19. If cosec q = 3x and cot q = 3 , then find the value of d x2 – 1 n .
x x2

20. What is the value of (1 + cot2 q)sin2 q ?

21. What is the value of sin2 q + 1 ?
1 + tan2 i

22. Write the value of sin A cos (90° – A) + cos A sin (90° – A).

23. What is the maximum value of 2 ? Justify your answer.
cosec i
QQ Short Answer Questions-II: [3 marks each]

24. If 4 tanq = 3, evaluate d 4 sin i – cos i + 1 n [CBSE 2018 (30/1)]
4 sin i + cos i – 1

268 Xam idea Mathematics–X

25. If tan 2A = cot(A – 18°), where 2A is an acute angle, find the value of A. [CBSE 2018 (30/1)]

26. Prove that: [CBSE 2018 (C)(30/1)]
[CBSE 2018 (C)(30/1)]
=11 ++ tcaont22 AA  =11++ tcaont AA 2 tan2 A [CBSE 2019 (30/1/1)]

27. Evaluate:

cos 58° + sin 22° – cos 38° cosec 52°
sin 32° cos 68° 3 (tan 18° tan 35° tan 60° tan 72° tan 55°)

28. Prove that (1 + cotA – cosecA)(1 + tanA + secA) = 2

29. Evaluate : [CBSE 2019(30/2/1)]

d 3 sin 43° 2 cos 37° cosec 53°
cos 47° tan 5° tan 25° tan 45° tan 65° tan 85°
n–

30. Find A and B if sin (A + 2B) = 3 and cos(A + 4B) = 0, where A and B are acute angles.
2
[CBSE 2019(30/2/2)]

31. Find the value of:

c 3 tan 41° 2 – d tan 10° tan sin 35° sec 55° 70° tan 80° 2 [CBSE 2019(30/2/3)]
cot 49° 20° tan 60° tan
m n

32. A, B and C are interior angles of a triangle ABC. Show that [CBSE 2019(30/3/1)]

(i) sin  B+C = cos A
 2  2
of tan d B +C
(ii) If A = 90°, then find the value 2 n.
B) = A+ B
33. If tan (A + B) = 1 and tan (A – 1 , 0° < < 90°, A > B, then find the value of A and B.
3 [CBSE 2019(30/3/1)]

34. Prove that:

(sinq + 1 + cos q)(sin q – 1 + cos q). sec q cosec q = 2 [CBSE 2019(30/4/2)]
[CBSE 2019(30/4/2)]
35. Prove that:
[CBSE 2019(C)(30/1/1)]
sec i – 1 + sec i +1 =2 cosec i
sec i +1 sec i –1 [CBSE 2019(C)(30/1/1)]

36. Prove that:

tan A – tan A = 2 cosec A
1 + sec A 1 – sec A

37. Prove that:

1 + 1 cot2 q =cos ecq
+ cos ecq
24
38. In DABC, right-angled at C, find cos A, tan A and cosec B if sin A = 25 .

39. In Fig. 10.11, find sin A, tan A and cot A.

40. If cot q = 1 , show that = 1 – cos2 i = 3 ..
3 2 – sin2 i 5

41. If tan q = 1 , find other five trigonometric ratios. Fig. 10.11
3
42. Write all the other trigonometric ratios of ∠B in terms of tan B.

43. In DOPQ, right-angled at P, OP = 7 cm, and OQ – PQ = 1 cm. Determine the values of sin Q and
cos Q.

Introduction to Trigonometry 269

44. In DABC, right-angled at B, AB = 3 cm and ∠BAC = 60°. Determine the lengths of the sides BC
and AC.

Evaluate the following: (Q. 45 to 53)
45. cos 90° sin 0° – sin 0° cos 90°.

46. cosesc6600° °–+cotatn454°5+° –cosisne3c 03°0°
47. 2sin2 30° – 3cos2 45° + tan2 60°.

48. tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos2 90° .
cosec 30° + sec 60°– cot2 30°

49. ctoasnec4350° ° + sec 60° – 3 sin 90° .
cot 45° 2 cos 0°

50. cosescec7200° ° + sec 59° .
cot 31°
51. tan 48° tan 23° tan 42° tan 67°.

52. csions1800°° + cos 59° cosec 31°.
53. cosec (65° + q) – sec (25° – q) – tan (55° – q) + cot (35° + q).

54. If sec 2A = cosec (A – 42°) where 2A is an acute angle, find the value of A.
55. If tan (A + B) = 3 and tan (A – B) = 0, 0° < A + B ≤ 90°, find sin (A + B) and cos (A – B).
56. If sin q + cos q = 3 , then prove that tan q + cot q = 1.
57. Given that a + b = 90°, show that cos a cosec b – cos a sin b = sin a .

58. If tan q = a , prove that a sin i – b cos i = a2 – b2 .
b a sin i + b cos i a2 + b2

59. If sec q = 5 , find the value of sin i – 2 cos i .
4 tan i – cot i

60. If A = 30° and B = 30°, verify that sin (A + B) = sin A cos B – cos A sin B.

61. If cot q = 15 , then evaluate (2 + 2 sin i) (1 – sin i) .
8 ]1 + cos ig(2 –2 cos i)

62. If sec q = x + 1 , prove that sec q tan q = 2x or 1 .
x 2x

63. If cosec q = 13 , find the value of 2 sin i – 3 cos i .
12 4 sin i – 9 cos i

64. If sin q = a2– b2 , find 1 + tan q cos q.
a2 + b2
65. Find the value of x if tan x = sin 45° cos 45° + sin 30°

Prove the following identities (Q. 66-69)

66. tan q + tan (90 – q) = sec q sec (90 – q)

67. tan4 q + tan2 q = sec4 q – sec2 q

68. cot q – tan q = 2 cos2 i –1 .
sin i cos i

69. (cosec q – cot q)2 = 1 − cos θ
1 + cos θ

270 Xam idea Mathematics–X

QQ Long Answer Questions: [5 marks each]

70. Prove that: [CBSE 2018 (30/1)]

sin A – 2 sin3 A = tan A .
2 cos3 A – cos A

71. If sin(A + 2B) = 3 and cos(A + 4B) = 0, A > B, and A + 4B ≤ 90°, then find A and B.
2
[CBSE 2018 (C)(30/1)]

72. Prove that: [CBSE 2019 (30/1/1)]

sin A – cos A + 1 = 1 . [CBSE 2019 (30/2/1)]
sin A + cos A –1 sec A – tan A
[CBSE 2019 (30/3/1)]
73. Prove that: [CBSE 2019 (30/3/3)]

sin i =2+ sin i .
cot i + cosec i cot i – cosec i
1
74. If 1+ sin2 q = 3 sin q cos q, then prove that tan q = 1 or tan q = 2 .

75. Prove that:

1 tan3 i + 1 cot3 i = sec i cosec i – 2 sin i cos i .
+ tan2 i + cot3 i

76. If sec q + tan q = m, show that m2 –1 = sin i . [CBSE 2019 (30/4/2)]
m2 +1

77. Prove that: [CBSE 2019 (30/4/3)]

2(sin6 q + cos6 q) – 3(sin4 q + cos4 q) + 1 =0 .

78. Prove that: [CBSE 2019 (30/5/1)]
+ cot i + tan i) (sin i –
(1 (sec3 i – cosec3 i) cos i) = sin2 i cos2 i .

79. If tan x = n tan y and sin x = m sin y, prove that cos2 x = m2 –1 . [CBSE 2019 (C) (30/1/1)]
n2 –1

80. If x sin3 q + y cos3 q = sin q cos q and x sin q = y cos q, prove that x2 + y2 = 1. [CBSE 2019 (C) (30/1/1)]

81. If a sin q + b cos q = c, then prove that a cos q – b sin q = a2 + b2 – c2 .

82. If a cos q + b sin q = m and a sin q – b cos q = n, prove that a2 + b2 = m2 + n2.
Prove the following identities (Q. 83 – 88)

83. sec2 i + cosec2 i = tan q + cot q
84. sin A(1 + tan A) + cos A (1 + cot A) = sec A + cosec A

85. ttaannii + sec i – 1 = 1 + sin i
– sec i +1 cos i

86. (1 + cot q + tan q)(sin q – cos q) = sec i – cosec i
cosec2 i sec2 i

87. 1s+incoθs(1θ+−csoins θ2 )θ = cot q

88. (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A.

89. If x = a sec q + b tan q and y = a tan q + b sec q, prove that x2 – y2 = a2 – b2.

Introduction to Trigonometry 271

Answers

1. (i) (c) (ii) (a) (iii) (b) (iv) (c) (v) (c)

2. (i) 0 (ii) 1 (iii) 3 (iv) 1 (v) 20°
3. 0 4. A = 38° 4

5. 1 6. 2 7. 4 8. sec a = 13
7 12

9. 1 10. cos 23° + sin 15° 11. 1 12. k = 1 13. 13
12

14. 1 15. 0 16. 0 18. 29° 19. 1 20. 1
9

21. 1 22. 1 23. 2 24. 13 25. 36° 27. 5
11 3

29. 8 30. A = 30°, B = 15° 31. 26 32. (ii) 1
3

33. A = 37.5°, B = 7.5° 3=8. cos A 2=75 , tan A 24 , cosec B= 25
7 7

39=. sin A 53=, tan A 3 , cot A = 4
4 3

41. sin θ = 1 , cos θ = 3 , cosec θ = 10, sec θ = 10 , cot θ = 3
10 10 3

42. sin B = tan B , cos B = 1 , cosec B = 1 + tan2 B , sec B = 1 + tan2 B, cot B = 1
1 + tan2 B 1 + tan2 B tan B tan
B

4=3. sin Q 2=75 , cos Q 24 44. BC = 3 3 cm, AC = 6 cm 45. 0 46. 3
25 5

47. 2 48. 9 49. 1 50. 2 51. 1 52. 2

53. 0 54. A = 44° 55. sin (A + B) = 3 , cos (A – B) = 1 59. 12
63. 3 2 7

61. 225 64. 2a2
64 a2 + b2

65. x = 45° 71. A = 30° B = 15°

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)
[NCERT Exemplar]
(i) If cos A = 4 , then the value of tan A is
5

(a) 3 (b) 3 (c) 4 (d) 5
5 4 3 3

272 Xam idea Mathematics–X

(ii) The value of the expression cosec (58° + i) – sec (32° – i) is
tan 45° + tan (45° + i) – cot (45° – i)

(a) 1 (b) 1 (c) 0 (d) 2
2

(iii) The value of tan2 60° – sin2 30° is
tan2 45° + cos2 30°

(a) 7 (b) 11 (c) 13 (d) 11
11 13 11 7
(iv) If sin A + sin2 A = 1, then the value of the expression cos2 A + cos4 A is

(a) 1 (b) 1 (c) 2 (d) 3
2

2. Fill in the blanks. (3 × 1 = 3)

(i) Reciprocal of tangent of an angle is _____________ .
(ii) If cos A + cos2A=1, then sin2A + sin4A = _____________ .

(iii) If a cos θ – b sin θ = c, then a sin θ + b cos θ = _____________ .

3. Solve the following questions. 3 (3 × 1 = 3)
4
(i) In a right angled triangle if tan i = . What is the measure of the greatest side of the
triangle?

(ii) Evaluate: sin237° + sin253° + sin290°.

(iii) In a right angled triangle if cos i = 1 , sin i = 3 , what is the value of tan θ?
2 2

Section B

QQ Solve the following questions. (3 × 2 = 6)
4. Prove that: sin6 θ + cos6 θ + 3 sin2 θ cos2 θ = 1 [Exemplar]
5. Prove that : (sin4 θ – cos4 θ + 1) cosec2 θ = 2
[NCERT Exemplar]
6. Show that :1 + 1 cot2 a a = cosec a
+ cosec (3 × 3 = 9)

QQ Solve the following questions. [NCERT]
3
7. If sin A = 4 , calculate cos A and tan A. (3 × 5 = 15)
[NCERT]
8. Show that cos 38° cos 52° – sin 38° sin 52° = 0. [NCERT]

9. Prove that : sin i – 2 sin3 i = tan i
2 cos3 i – cos i

QQ Solve the following questions.

10. Prove : tan i + 1 cot i i = 1 + sec i cosec i
1 – cot i – tan

11. Without using trigonometric tables, evaluate the following:

32 cosec2 58° – 2 cot 58° tan 32° – 5 tan 13° tan 37° tan 45° tan 53° tan 77°
3 3

Introduction to Trigonometry 273

12. Prove that: cos A – sin A + 1 = cosec A + cot A [NCERT]
cos A + sin A –1 [NCERT]

OR zzz

Prove that : 1 + sin A = sec A + tan A
1 – sin A

Answers

1. (i) (b) (ii) (c) (iii) (d) (iv) (a)

2. (i) cotangent (ii) 1 (iii) ! a2 + b2 – c2

3. (i) 5 units (ii) 2 (iii) 3

7. cos A = 7 , tan A = 3 11. –1
4 7



274 Xam idea Mathematics–X

Heights and 11
Distances

BASIC CONCEPTS – A FLOW CHART

Heights and Distances 275

MORE POINTS TO REMEMBER Fig. 11.1

 In solving problems observer is represented by a point if his
height is not given.

 In solving problems object is represented by a line segement
and some times by a point if height or length is not
considered.

For example, AB is tower and point C is observer.
 A line drawn parallel to earth surface is called horizontal

line.
 The angle of elevation of a point C as seen from a point A is

always equal to the angle of depression of A as seen from C.

Fig. 11.2

Thus, angle of elevation = angle of depression
 The angle of elevation and depression are always acute

angles.
 If the observer moves towards the objects like tower,

building, cliff, etc. then angle of elevation increases and
if the observer moves away from the object, the angle of
elevation decreases.
 If the angle of elevation of sun decreases, then the
length of shadow of an object increases and vice-versa.

Fig. 11.3

Fig. 11.4

 If in problems, the angle of elevation of an object is given, then we conclude that the object
is at higher altitude than observer. The angle of depression implies that observer is at
higher altitude than object.

276 Xam idea Mathematics–X

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. A pole 6 m high casts a shadow 2 3 m long on the ground, then the Sun’s elevation is

[NCERT Exemplar]

(a) 60° (b) 45° (c) 30° (d) 90°

2. Which are the angles of depression from the observing positions A and C respectively of the
object at E in Fig. 11.5?

CA

45°

60° D B
E

Fig. 11.5

(a) 30°, 60° (b) 45°, 30° (c) 45°, 60° (d) 15°, 30°

3. If the altitude of the Sun is at 60°, then the height of the vertical tower that will cast a shadow

of length 20 m is

(a) 20 3 m (b) 20 m (c) 15 m (d) 40 m
3 3 3

4. The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is

(a) 30° (b) 45° (c) 60° (d) 90°

5. A ladder 18 m long makes an angle of 60° with a wall. The height of the point where the ladder
reaches the wall is

(a) 9 3 m (b) 18 3 m (c) 18 m (d) 9 m

6. The length of the ladder making an angle of 45° with a wall and whose foot is 7 m away from

the wall is

(a) 72 m (b) 7 2 m (c) 14 2 m (d) 14 m
2

7. The angle formed by the line of sight with the horizontal when the object viewed is below the

horizontal level is

(a) corresponding angle (b) angle of elevation

(c) angle of depression (d) none of these

8. If the angle of elevation of top of a tower from a point on the ground which is 20 3 m away
from the foot of the tower is 30°, then the height of the tower is

(a) 60 m (b) 30 m (c) 25 m (d) 20 m

9. The height of a pole is 10 m. What is the length of the shadow when Sun’s altitude is 30°?

(a) 10 m (b) 10 3 m (c) 10 m (d) 15 m
3

10. If the angles of elevation of the top of a tower from the two points at a distance of 2 m and 8 m
from the base of the tower and in the same straight line with it are complementary, then the
height of the tower is

(a) 3 m (b) 4 m (c) 5 m (d) 6 m

Heights and Distances 277

11. A 1.6 m tall girl stands at distance of 3.2 m from a lamp post and casts shadow of 4.8 m on the

ground, then the height of the lamp post is 8
3
(a) 8 m (b) 4 m (c) 6 m (d) m

12. A tower casts a shadow 90 m long and at the same time another tower casts a shadow of 120 m

on the ground. If the height of the second tower is 80 m, then the height of the first tower is

(a) 60 m (b) 5.5 m (c) 50 m (d) 40 m

13. An observer 1.6 m tall is 20 3 m away from a tower. The angle of elevation from his eye to the
top of the tower is 30°. The height of the tower is

(a) 21.6 m (b) 23.2 m (c) 24.72 m (d) None

14. If a 30 m ladder is placed against a 15 m wall such that it just reaches the top of the wall, then
the elevation of the wall is equal to

(a) 45° (b) 30° (c) 60° (d) 50°

15. A boy is standing at the top of the tower and another boy is at the ground at some distance
from the foot of the tower, then the angle of elevation and depression between the boys when
both boys look at each other will be

(a) equal (b) angle of elevation will be greater

(c) cannot be predicted (d) angle of depression will be greater

Answers

1. (a) 2. (c) 3. (a) 4. (b) 5. (a) 6. (b)
7. (c) 8. (d) 9. (b) 10. (b) 11. (d) 12. (a)

13. (a) 14. (b) 15. (a)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. _______________ is the line drawn from the eye of an observer to the point in the object viewed
by the observer .

2. The _______________ of an object viewed, is the angle formed by the line of sight with the
horizontal when it is above the horizontal level, i.e; the case when we raise our head to look at
the object.

3. The _______________ of an object viewed, is the angle formed by the line of sight with the
horizontal when it is below the horizontal level i.e: the case when we lower our head to look at
the object.

4. __________ are used to find height or length of an object or distance between two distant objects.

5. When the length of the shadow of a pole is equal to its height, then the angle of elevation of
source of light is _______________ .

6. For two towers of height h1 and h2; h1 = tan 60° and h2 = tan 30°, at mid points of line joining
x x
their feet then h1 : h2 = _______________.

7. The angle of elevation of the top of a tower is 30°, if the height of the tower is doubled, then the

angle of elevation of its top will be _______________ .

8. The angle of elevation and angle of depression are always _______________ angles.

9. As the observer moves towards the object the angle of elevation _______________ .

10. _______________ is a tool used to measure height of tall things, that you can't possibly reach to
the top of the object.

278 Xam idea Mathematics–X

Answers

1. Line of sight 2. angle of elevation 3. angle of depression

4. Trigonometric ratios 5. 45° 6. 3 : 1 7. same 8. acute

9. increases 10. Clinometer

Very Short Answer Questions [1 mark]

1. If a pole 12 m high casts a shadow 4 3 m long on the ground, find the Sun’s elevation.
[NCERT Exemplar]

Sol. tan q = 12 = 3= 3
43 3

⇒ q = 60° 12 m

4√3 m

Fig. 11.6

2. An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation
of the top of the tower from the eye of the observer.

Sol. PQ = MB = 1.5m

AM = AB – MB = 22 – 1.5 = 20.5m

Now in DAPM

tan q = AM = 20.5 = 1 ⇒ tanθ = tan 45°
PM 20.5

⇒ θ = 45°

Fig. 11.7

3. A ladder 15 m long makes an angle of 60° with the wall. Find the height of the point where the

ladder touches the wall. [CBSE (F) 2017]

Sol. cos 60° = x
15

⇒ 1 = x
2 15

⇒ x= 15 m = 7.5 m Fig. 11.8
2

4. The ratio of the height of a tower and the length of its shadow on the ground

is 3 : 1. What is the angle of elevation of the sun? [CBSE Delhi 2017]

Sol. Given: AB = 3
BC 1

Then, tan q = AB = 3
BC

⇒ q = 60° Fig. 11.9

Heights and Distances 279

5. If the angle of elevation of a tower from a distance of 100 m from its foot is 60°, then what will

be the height of the tower? [NCERT Exemplar]

Sol. Let h be the height of the tower.

tan 60° = AB
BC

3 = h
100

h = 100 3 m

Fig. 11.10

6. In Fig. 11.11, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60°
to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the

length of the ladder. (Use 3 =1.73) [CBSE Delhi 2016]

Sol. DB = (6 – 2.54)m = 3.46 m

In DBDC, sin 60° = BD
CD

⇒ 3 = 3.46 ⇒ CD = 3.46 × 2
2 CD 1.73
Fig. 11.11

\ DC = 4 m

7. An observer, 1.7 m tall, is 20 3 m away from a tower. The angle of elevation from the eye of

observer to the top of tower is 30°. Find the height of tower. [CBSE (F) 2016]

Sol. Let AB be the height of tower and DE be the height of observer.

Then in DACD, AC = tan 30°.
DC

⇒ x 3 = tan 30° = 1 ⇒ x = 20 m (Fig. 11.12)
20 3

\ AB = 20 + 1.7 = 21.7 m

Fig. 11.12

8. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of

elevation of the sun ?           [CBSE (AI) 2017]

Sol. In DABC B
30 m
tan θ = AB = 30 = 3
AC 10 3

⇒ tan θ = tan 60°

⇒ θ = 60°

θ
A 10 m C

Fig. 11.13

280 Xam idea Mathematics–X

Short Answer Questions-I [2 marks]

1. Find the angle of elevation of the sun when the shadow of a pole h m high is 3 h m long.

Sol. In DABC

tan q = AC = h
BC 3h

⇒ tan q = 1 = tan 30°
3

\ q = 30°

Fig. 11.14

2. The height of a tower is 12 m. What is the length of its shadow when Sun’s altitude is 45°?

Sol. Let AB be the tower [Fig. 11.15].

Then, ∠C = 45°, AB = 12 m

tan 45° = AB = 12 12 ⇒ BC = 12 m
BC BC ⇒ 1 = BC

\ The length of the shadow is 12 m.

Fig. 11.15

3. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top
of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with

the ground level is 30° [NCERT]

Sol. Let AB be the vertical pole and AC be the long rope tied to point C.

In right DABC, we have [Fig. 11.16]

sin 30° = AB   ⇒ 1 = AB ⇒ 20 = AB
AC 2 20 2

⇒ AB = 10 m Fig. 11.16
Therefore, height of the pole is 10 m.

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away
from the foot of the tower, is 30°. Find the height of the tower.

Sol. Let BC be the tower whose height is h metres and A be the point at a distance of 30 m from the
foot of the tower. The angle of elevation of the top of the tower from point A is given to be 30°.

Now, in right angle DCBA we have,

tan 30° = BC = h ⇒ 1 = h
AB 30 3 30

⇒ h = 303 = 30 × 3 = 30 3 = 10 3m
3 3 3

Hence, the height of the tower is 10 3 m. Fig. 11.17

Short Answer Questions-II [3 marks]

1. A tree breaks due to storm and the broken part bends, so that the top of the tree touches the

ground making an angle 30° with it. The distance between the foot of the tree to the point

where the top touches the ground is 8 m. Find the height of the tree. [NCERT]

Heights and Distances 281

Sol. In right angle DABC, AC is the broken part of the tree (Fig. 11.18).

So, the total height of tree = (AB + AC)

Now in right angle DABC ,

tan 30° =  AB   ⇒  1 = AB ⇒  AB = 8
BC 3 8  3

Again, cos 30°   =  BC
AC

⇒ 3 = 8 ⇒ AC = 16
2 AC 3
Fig. 11.18
Hence, the height of the tree = AB + AC

= 8 + 16 = 24 = 24 × 3 = 24 3 = 8 3m
3 3 3 3 3 3

2. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from

the base of the tower and in the same straight line with it are complementary. Prove that the

height of the tower is 6 m. [NCERT]

Sol. Let OA be the tower of height h metre and P, Q be the two points at distance of 9 m and 4 m
respectively from the base of the tower.

Now, we have

OP = 9 m, OQ = 4 m

Let ∠APO = q, ∠AQO = (90° – q)

and OA = h metre (Fig. 11.19)

Now, in DPOA, we have

tan q = OA = h ⇒ tan q = h ... (i)
OP 9 9
Again, in DAQO we have
Fig. 11.19

tan (90° – q) = OA = h ⇒ cot q = h ... (ii)
OQ 4 4

Multiplying (i) and (ii), we have

tan q × cot q = h × h ⇒ 1= h2 ⇒ h2 = 36
9 4 36

h=±6

Height cannot be negative.

Hence, the height of the tower is 6 metre. Hence proved

3. Determine the height of a mountain if the elevation of its top at an unknown distance from
the base is 30° and at a distance 10 km further off from the mountain, along the same line, the
angle of elevation is 15°. (Use tan 15° = 0.27)

Sol. Let AB be the mountain of height h kilometres. Let C be a point at a distance of x km, from the
base of the mountain such that the angle of elevation of the top at C is 30°. Let D be a point at a
distance of 10 km from C such that angle of elevation at D is of 15°.

In DABC (Fig. 11.20), we have

tan 30° = AB ⇒ 1 = h
AC 3 x

⇒ x = 3h Fig. 11.20

282 Xam idea Mathematics–X

In DADB, we have

tan 15° = AB ⇒ 0.27 = x h
AD + 10

⇒ 0.27 (x + 10) = h ...(i)

Substituting x = 3h in equation (i), we get

0.27 ( 3h + 10) = h

⇒ 0.27 × 3h + 0.27 × 10 = h

⇒ 2.7 = h – 0.27 × 3h ⇒ 2.7 = h (1 – 0.27 × 3 )

⇒ 2.7 = h (1 – 0.46) ⇒ =h 0=2.5.74 5

Hence, the height of the mountain is 5 km.

4. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s
altitude is 30° than when it is 60°. Find the height of the tower.

Sol. In Fig. 11.21, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°,
i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the
length of the shadow, when the angle of elevation is 30°.

Now, let AB be h m and BC be x m.

According to the question, DB is 40 m longer than BC.

So, BD = (40 + x) m

Now, we have two right triangles ABC and ABD.

In DABC, tan 60° = AB or 3 = h
BC x

⇒ x 3 = h ...(i)
In DABD,
tan 30° = AB Fig. 11.21
BD

i.e., 1 = h ...(ii)
3 x + 40

Using (i) in (ii), we get (x 3 ) 3 = x + 40, i.e., 3x = x + 40

i.e., x = 20

So, h = 20 3 (From (i))

Therefore, the height of the tower is 20 3 m.

5. From a point P on the ground, the angle of elevation of the top of a 10m tall building is 30°. A
flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff
from P is 45°. Find the length of the flagstaff and the distance of the building from the point
P. (You may take 3 = 1.732).

Sol. In Fig. 11.22, AB denotes the height of the building, BD the flagstaff
and P the given point. Note that there are two right triangles PAB and

PAD. We are required to find the length of the flagstaff, i.e., BD and
the distance of the building from the point P, i.e., PA.

Since, we know the height of the building AB, we will first consider the

right DPAB.

We have, AB ⇒ 1 = 10 Fig. 11.22
tan 30° = AP 3 AP

Heights and Distances 283

⇒ AP = 10 3

i.e., the distance of the building from P is 10 3 m = 10 × 1.732 = 17.32 m.

Next, let us suppose DB = x m. Then, AD = (10 + x) m.

Now, in right DPAD,

tan 45° = AD = 10 + x ⇒ 1 = 10 + x ⇒ 10 3 = 10 + x
AP 10 3 10 3

i.e., x = 10( 3 – 1) = 7.32

So, the length of the flagstaff is 7.32 m.

6. A contractor plans to install two slides for the children to play in a park. For the children

below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is

inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep

slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the

length of the slide in each case? [NCERT]

Sol. Let AC be a steep slide for elder children and DE be a slide for
younger children. Then AB = 3 m and DB = 1.5 m (Fig. 11.23).

Now, in right angle DDBE, we have

sin 30° = BD = 1.5
DE DE

⇒ 1 = 1.5 \ DE = 2 × 1.5 = 3 m
\ 2 DE
Fig. 11.23
Length of slide for younger children = 3 m

Again, in right angle DABC, we have

  sin 60° = AB ⇒ 3 = 3
AC 2 AC

⇒ AC = 6 × 3 = 63 = 2 3m
3 3 3

So, the length of slide for elder children is 2 3 m.

7. A kite is flying at a height of 60 m above the ground. The string attached to the kite is

temporarily tied to a point on the ground. The inclination of the string with the ground is 60°.

Find the length of the string, assuming that there is no slack in the string. [NCERT]

Sol. Let AB be the horizontal ground and K be the position of the kite and its height from the ground
is 60 m and let length of string AK be x m. (Fig. 11.24)

∴ ∠KAB = 60°

Now, in right angle DABK we have

sin 60° = BK = 60 ⇒ 3 = 60 ⇒ 3x = 120
AK x 2 x

\ x = 120 # 3 = 120 3 = 40 3m
3 3 3

So, the length of string is 40 3 m. Fig. 11.24

8. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation

from his eyes to the top of the building increases from 30° to 60° as he walks towards the

building. Find the distance he walked towards the building. [NCERT]

284 Xam idea Mathematics–X

Sol. Let AB be the building and PQ be the initial position of the boy (Fig. 11.25) such that

∠APR = 30° and AB = 30 m

Now, let the new position of the boy be P'Q' at a distance QQ'.

Here, ∠AP'R = 60°

Now, in DARP, we have

tan 30° = AR = AB − RB
PR PR

⇒ 1 = 30 − 1.5 = 28.5
3 PR PR

PR = 28.5 × 3

Again, in DARP' we have

tan 60° = AR ⇒ 3 = 28.5 Fig. 11.25
P'R P'R

    P'R = 28.5 × =33 28.5 =3 9.5 3
3 3

Therefore, required distance, QQ' = PP' = PR – P'R = 28.5 3 – 9.5 3 = 19 3

Hence, distance walked by the boy is 19 3 m.

9. From a point on a bridge across a river, the angles of depression of the banks on opposite sides
of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, find
the width of the river.

Sol. In Fig. 11.26, A and B represent points on the bank on opposite sides of the river, so that AB

is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3m. We are

interested to determine the width of the river, which is the length of the side AB of the DAPB.

In right DADP, ∠A = 30°

So, PD
tan 30° = AD

i.e., 13 = A3D or AD = 3 3 m

Also, in right DPDB, Fig. 11.26

PD = tan 45° ⇒ 3 = 1
DB DB

\ DB = 3m
Now, AB = BD + AD = 3 + 3 3 = 3 (1 + 3 ) m

Therefore, the width of the river is 3( 3 + 1) m.

10. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression
of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the
lighthouse, find the distance between the two ships. [NCERT]

Sol. Let AB be the lighthouse of height 75 m and P, Q be the
position of the two ships whose angles of depression are 45°
and 30°, respectively (Fig. 11.27).

Let BP = x m and PQ = y m, we have

∠APB = 45° and ∠AQB = 30°

Now, in DABP we have

tan 45° = AB ⇒ 1= 75 Fig. 11.27
BP x

Heights and Distances 285

⇒ x = 75 m ...(i)

Again, in DABQ we have

tan 30° = AB
BQ

⇒ 1 75 y ⇒ x + y = 75 3 ...(ii)
3 = x +

From (i) and (ii), we have

75 + y = 75 3

y = 75 3 – 75 ⇒ y = 75( 3 – 1)

Hence, the distance between two ships is 75( 3 – 1) metres.

11. Two ships are there in the sea on either side of a light house in such a way that the ships and

the light house are in the same straight line. The angles of depression of two ships as observed

from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find

the distance between the two ships. [Use 3 = 1.73] [CBSE Delhi 2014]

Sol. Let the distance between the two ships be d metres.

Let the distance of one ship from the light house be x metres. Then, the distance of the other ship
from the light house will be (d – x) metres.

In DACO, OC 200
AC x
tan 45° = =

200 ...(i)
          1 = x   ⇒    x = 200 m

In DBCO, tan 60° = OC
BC

200 d–x= 200
3 = d − x    ⇒   3

Substituting the value of x from (i)

200 Fig. 11.28
3
d – 200 =

d = 200 + 200 = 200  1 + 1
3  3

= 200  1 + 3  = 200 × (1 + 3 3) × 3
  3
3

= 200 ( 3 + 3) = 200 × (1.73 + 3) = 200 × 4.73 = 946 = 315.33 m
3 3 3 3

12. The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of

30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height

of 3000 3 m, find the speed of the aeroplane. [CBSE (AI) 2014]

Sol. Let A be point of observation and P and Q be positions of the plane. Let ABC be the line through
A and it is given that angles of elevation from point A to two positions P and Q are 60° and 30°.

286 Xam idea Mathematics–X

\ ∠PAB = 60°, ∠QAB = 30°

Height = 3000 3 m

So, in DABP, we have BP
tan 60° = AB

3 = 300A0B 3   ⇒  AB = 3000 m

In DACQ

tan 30° = CQ Fig. 11.29
AC

1 = 300A0C 3
3

AC = 9000 m

\ Distance = BC = AC – AB = 9000 m – 3000 m = 6000 m

Also, plane travels for 30 seconds.

Hence, speed of plane = 6000 = 200 m/s = 200 × 18 km/h = 720 km/h.
30 5

13. From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower

are 45° and 60° respectively. Find the height of the tower. [Take 3 = 1.73] [CBSE (AI) 2014]

Sol. Let the height of the building be AE = 60 m, the height of the tower is
‘h’. The distance between the base of the building and the tower be ‘d’.

In DADE, AE 60
DE d
tan 60° = ⇒ 3 =

d == 603 6=03 3 20 3

BC = 20 3 = 20 × 1.73 = 34.60 m

In DABC, tan 45° = AC ⇒ 1= AC
BC 34.60

⇒ AC = 34.60 m Fig. 11.30

Now, height of tower = AE – AC = 60 – 34.60 = 25.4 m

14. A vertical tower stands on a horizontal plane and is

surmounted by a flagstaff of height 5 m. From a point on

the ground the angles of elevation of the top and bottom of

the flagstaff are 60° and 30° respectively. Find the height

of the tower and the distance of the point from the tower.

(Take 3 = 1.732) [CBSE (F) 2016]

Sol. Let height of tower be x m and distance of

point from tower be y m.

(i) From the Fig. 11.31, x = tan 30° = 1 ⇒y= 3x
y 3

(ii) x+5 = tan 60° = 3 or x + 5 = 3 ( y = 3x )
y 3x

Fig. 11.31

Heights and Distances 287

⇒ x + 5 = 3x ⇒ x= 5 = 2.5
2

Height of tower = 2.5 m
Distance of point from tower = y = 3x

= (2.5 × 1.732) or 4.33 m

15. Two men on either side of a 75 m high building and in line with base of building observe the

angle of elevation of the top of the building as 30° and 60°. Find the distance between the two

men. (Use 3 = 1.73) [CBSE (F) 2016]

Sol. Let AB be the building having height 75 m and the angles of elevation are 30° and 60° from the

point M1 and M2 respectively;

In DABM1, AB = tan 30° = 1 ⇒  BM1 = 75 3 m
BM1 3

In DABM2, BBAMMB22 == ta7n53 60° = 3
= 25 3m

⇒

∴ M1M2 = M1B + BM2 Fig. 11.32

= 75 3 + 25 3 = 100 3 m = 173 m

\ Distance between two men = 173 m.

16. A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The

angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat

in m/h. [CBSE Delhi 2017]

Sol. Let the speed of boat be x m/min

\ CD = 2x

In DABC,

150 = tan 60° & y = 150 = 50 3m 150 m
y 3

In DABD

150 = tan 45° & 150 = 50 3 + 2x
y + 2x ( y = 50 3 m)
y 2x
Fig. 11.33
⇒ x = 25(3 – 3 )

Speed = 25(3 – 3 ) m/min

\ = 25 × 60 (3 – 3 ) m/h = 1500 (3 – 3 ) m/h

Long Answer Questions [5 marks]

1. From a point on the ground, the angles of elevation of the bottom and the top of a transmission

tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of

the tower. [NCERT]

Sol. Let AB be a building of height 20 m and BC be the transmission tower of height x m and D be
any point on the ground (Fig. 11.34).

Here, ∠BDA = 45° and ∠ADC = 60°

Now, in DADC, we have

288 Xam idea Mathematics–X

tan 60° = AC ⇒ 3 = x + 20
AD AD

⇒A⇒g a in1,=inAA2D0DDA =D B,xw+e32h0⇒a v e AtaDn =452°0 =m AADB ...(i)
Putting the value of AD in equation (i), we have ...(ii)

⇒ 20 = x + 20 ⇒ 20 3 = x + 20 Fig. 11.34
3

⇒ x = 20 3 – 20 = 20 ( 3 – 1) = 20 (1.732 – 1) = 20 × 0.732 = 14.64 m

Hence, the height of tower is 14.64 m.

2. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of

elevation of the top of the statue is 60° and from the same point, the angle of elevation of the

top of the pedestal is 45°. Find the height of the pedestal. [NCERT]

Sol. Let AB be the pedestal of height h metres and BC be the statue of height 1.6 m.

Let D be any point on the ground such that,

∠BDA = 45° and ∠CDA = 60°

Now, in DBDA, we have h
DA
tan 45° = AB = h ⇒ 1 =
\ DA DA

DA = h ...(i)

Again in DADC, we have

tan 60° = AC = AB + BC
AD AD

⇒ 3 = h +1.6 (From equation (i)) Fig. 11.35
h

⇒ 3 h = h + 1.6 ⇒ ( 3 – 1) h = 1.6

\ h= 1.6 = 1.6 × 3 + 1 = 1.6 ( 3 + 1) = 1.6 ( 3 + 1) = 0.8 × ( 3 + 1) m
3 −1 3− 3 + 1 3 −1 2
1

Hence, height of the pedestal is 0.8 ( 3 + 1) m.
3. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and

the angle of depression of its foot is 45°. Determine the height of the tower.
[NCERT, CBSE Delhi 2017]

OR
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the

angle of depression of its foot is 45°. Find the height of the tower. [Use 3 = 1·732]

[CBSE (F) 2017]

Sol. Let PQ be the building of height 7 metres and AB be the cable tower. Now it is given that the
angle of elevation of the top A of the tower observed from the top P of building is 60° and the
angle of depression of the base B of the tower observed from P is 45°. (Fig. 11.36)

So, ∠APR = 60° and ∠QBP = 45°

Let QB = x m, AR = h m then, PR = x m

Now, in DAPR, we have

Heights and Distances 289

tan 60° = AR ⇒ 3= h
PR x

⇒ 3 x = h ⇒ h = 3 x ...(i)

Again, in DPBQ we have

tan 45° = Pt Q ⇒ 1 = 7    ⇒ x = 7 ...(ii)
QB x

Putting the value of x in equation (i), we have

h= 3 ×7=7 3

i.e., AR = 7 3 metres

So, the height of tower = AB = AR + RB = 7 3 + 7 = 7( 3 + 1) m. Fig. 11.36

4. At a point, the angle of elevation of a tower is such that its tangent is 5 . On walking 240 m
12
3
nearer to the tower, the tangent of the angle of elevation becomes 4 . Find the height of the
tower.

Sol. In the Fig. 11.37 let AB be the tower, C and D be the positions of observation from where given

that tan f = 5 … (i) and tan q = 3 …(ii)
12 4

Let BC = x m, AB = y m

Now in right-angled triangle ABC,

(ii)txaan=ndq43=(yii i)xy, we ge t 43 = x y ...(iii) Fig. 11.37
From ...(iv)
⇒

Also in right-angled triangle ABD, we get

tan f = x + y ...(v)
240

From (i) and (v), we get

5 = x + y ⇒ 12y = 5x + 1200 ...(vi)
12 240

⇒ 12y = 5 × 4 y + 1200 (Using (iv))
3

⇒ 12 y − 20 y = 1200 ⇒ 36 y − 20 y = 1200
3 3

⇒ 16y = 3600 ⇒= y 3=16600 225

Hence, the height of the tower is 225 metres.

5. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m

from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant

is 60°. After some time, the angle of elevation reduces to 30° (Fig. 11.38). Find the distance

travelled by the balloon during the interval. [NCERT]

Sol. Let A and B be two positions of the balloon
and G be the point of observation. (eyes of the girl)

290 Xam idea Mathematics–X

Now, we have

AC = BD = BQ – DQ = 88.2 m – 1.2 m = 87 m

∠AGC = 60°, ∠BGD = 30°

Now, in DAGC, we have

tan 60° = AC
GC

⇒ 3 = 87
GC

⇒ CG = 87 = 87 × 3 = 87 × 3
3 3 3 3

⇒ GC = 29 × 3 …(i) Fig. 11.38
…(ii)
Again, in DBGD, we have 1 87
3 GD
tan 30° = BD   ⇒ =
GD

⇒ GD = 87 × 3

From (i) and (ii), we have

  CD = 87 × 3 – 29 × 3

= 3 (87 – 29) = 58 3

Hence, the balloon travels 58 3 metres.

6. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes
a car at an angle of depression of 30°, which is approaching the foot of the tower with a
uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find
the time taken by the car to reach the foot of the tower from this point.

[NCERT, CBSE Delhi 2017 (C)]

Sol. Let OA be the tower of height h, and P be the initial position of the car when the angle of
depression is 30°.

After 6 seconds, the car reaches to Q such that the angle of depression at Q is 60°. Let the speed
of the car be v metre per second. Then,

PQ = 6v ( Distance = speed × time)

and let the car take t seconds to reach the tower OA from Q (Fig. 11.39). Then, OQ = vt metres.

Now, in DAQO, we have

OA
tan 60° = QO

⇒ 3 = h ⇒ h = 3 vt ...(i)
vt

Now, in DAPO, we have

tan 30° = OA Fig. 11.39
PO

⇒ 1 = 6v h vt ⇒ 3 h = 6v + vt ...(ii)
3 +

Now, substituting the value of h from (i) into (ii), we have

3 × 3 vt = 6v + vt

Heights and Distances 291

⇒ 3vt = 6v + vt ⇒ 2vt = 6v ⇒ =t 62=vv 3

Hence, the car will reach the tower from Q in 3 seconds.

7. From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower
and in same straight line with its base, with angles of depression 30° and 45°. Find the distance

between the cars. [Take 3 = 1.732] [CBSE (AI) 2017]

Sol. Let AO be the tower of height 100 m. Car B and Car C are in opposite direction and at distance
of x m and y m respectively.

In DABO, 45° 30°

100 = tan 45° = 1
x

⇒ x = 100 m ...(i)
...(ii)
In DACO, 100m

100 = tan 30° = 1 45° 30°
y 3 xO y

⇒ y = 100 3 m

Distance between the cars = x + y Fig. 11.40

= 100 + 100 3 [From equation (i) and (ii)]

= 100 (1 + 3 )

= 100 (1 + 1.732) = 273.2 m

8. Two poles of equal heights are standing opposite to each other on either side of the road,

which is 80 m wide. From a point between them on the road, the angles of elevation of the top

of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the

point from the poles. [NCERT; CBSE 2019 (30/1/2)]

Sol. Let AB and CD be two poles of equal height h metre and let
P be any point between the poles, such that

∠APB = 60° and ∠DPC = 30°

The distance between two poles is 80m. (Given)

Let AP = x m, then PC = (80 – x) m

Now, in DAPB, we have

tan 60° = AB = h Fig. 11.41
AP x

⇒ 3 = h ⇒ h= 3 x ...(i)
x

Again in DCPD, we have

tan 30° = DC = h x)
PC (80 −

⇒ 1 = h ⇒ h = 80 − x ...(ii)
3 80 − 3
x

From (i) and (ii), we have

3x = 80 − x ⇒ 3x = 80 – x
⇒ 3
⇒ =x 8=40 20 m
4x = 80

292 Xam idea Mathematics–X