Shubharambha OPT Mathematics 10 Final for CTP 2077

= A × A – 1
cot 2 cot 2

cot A + cot A
2 2

A
cot2 2 –1
\ CotA =
A
2cot 2

Trigonometric Ratios of angle A in terms of A
1. sinA = sin 3
2A A
3 + 3

= sin23A . cosA3 + cos23A . sin A
3

= 2sinA3 . cosA3 . cosA3 + 1 – 2sin2 A . sin A
3 3

= 2sinA3 . cos2 A + sinA3 – 2sin3 A
3 3

= 2sinA3 1 – sin2 A + sinA3 – 2sin3 A
3 3

= 2sinA3 – 2sin3 A + sinA3 – 2sin3 A
3 3

∴ sinA = 3sinA3 – 4sin3 A
3

2. cosA = cos 2A + A
3 3

= cos23A . cosA3 + sin23A . sinA3

= 2cos2 A – 1 . cosA3 – 2 sinA3 . cosA3 × sin A
3 3

= 2cos3 A – cos A – 2cosA3 1 – cos2 A
3 3 3

= 2cos3 A – cos A – 2cosA3 + cos3 A
3 3 3

= 4cos3 A – 3cosA3
3

∴ cosA = 4cos3 A – 3cos3 A
3 3

Similarly, A A
3 3
3. tanA = 3tan A – tan3 A 4. cotA = cot3 – 3 cot
3 3
A
1 – tan2 A 3cot2 3 –1
3

Infinity Optional Mathematics Book - 10 201

Comparative study of trigonometric ratios of multiple and sub-multiple

angles.

Multiple Angle Sub-multiple angle

1. sin2A = 2sinA . cosA 1. sinA = 2sinA2 .AcosA2

2. sin2A = 2tanA 2. sinA = 2tan 2
+ tan2A
1 1 + tan2 A
A 2

3. sin2A = 2cotA 3. sinA = 2cot 2
+ cot2A
1 1 + cot2 A
2

4. cos2A = cos2A – sin2A 4. cosA = cos2A2 – sin2 A
5. cos2A = 2cos2A – 1 2
6. cos2A = 1 – 2sin2A
5. cosA = 2cos2A2 – 1
6. cosA = 1 – 2sin2A2

A
1 – tan2 2
7. cos2A = 1 – tan2A 7. cosA = 1 +
1 + tan2A A
tan2 2

A
cot2 2 –1
8. cos2A = cot2A – 1 8. cosA = +1
cot2A + 1 cot2 A
2
9. 2cos2A2 = 1 + cosA
9. 2cos2A = 1 + cos2A

10. 2sin2A = 1 – cos2A 10. 2sin2 A = 1 – cosA
2 A

11. tan2A = 1 2tanA 11. tanA = 2tan 2 A
– tan2A – tan2 2
1
cot2A – 1
12. cot2A = 2cotA cot2 A – 1
2 A
12. cotA =
13. sin3A = 3sinA – 4sin3A
2cot 2
14. cos3A = 4cos3A – 3cosA 3sinA3 A
13. sinA = – 4sin3 3

15. tan3A = 3tanA – tan3A 14. cosA = 4cos3A3 – 3cosA3 A
1 – 3tan2A A 3
3 tan 3 – tan3
15. tanA =
A
16. cot3A = cot3A – 3cotA 1 – 3tan2 3
3cot2A – 1
cot3 A – 3cot A
16. cotA = 3 3
A
3cot2 3 – 1

202 Infinity Optional Mathematics Book - 10

WORKED OUT EXAMPLES

1. Find the value of sin q, cos q and tan q when sin θ = 4
Solution : 2 5

Here, sin θ = 4
2 5

cos θ = 1 – sin2 θ = 1– 4 2 3
2 2 5 5
=

Now, sinq = 2sin θ . cos θ
2 2

= 2 × 4 × 3 = 24
5 5 25
θ
cosq = 1 – 2sin2 2

= 1 – 2 4 2
5

= 1 – 32
25
–7
= 25

sinθ 24 24
cosθ –7
tanq = = 25 =
–7

25

2. If cos30°­= 3 show that sin15° = 3 – 1
2 22
Solution:

Here, cos30º = 3
2
1 – 2sin2320° = θ
or, 3 [∴ cosq = 1 – 2sin2 2 ]
2
3
or, 2sin215º = 1 – 2

or, sin215º = 2 – 3
4

or, sin15º = ± 2– 3 × 2
4 2

or, sin15º = ± 4–2 3 = 3+1–2 3
8 8

or, sin15º = ± ( 3 )2 – 2 . 3 . 1 + 12
8

Infinity Optional Mathematics Book - 10 203

or, sin15º = ± ( 3 – 1)2
8

or, sin15º = ± 3 –1
22
Since, 15º lies in first quadrant where the value of sin15º is +ve

∴ sin15º = 3 –1
22

3. If cos315º = 1 , show that cos157 1° = ­–21 2+ 2
Solution : 2 2

Here, Cos315º = 1 θ
2 2
315°
or, 2cos2 2 – 1 = 1 ∴ cosθ = 2cos2 – 1
2

or, 2cos2 315° = 1 + 1
2 2

or, cos2 315° = 2+ 1
2 22

or, cos2 315° = 2+ 1 × 2
2 22 2

or, cos 315° = ± 2+ 2
2 4

or, cos 315° = ± 2 + 2
2 2

Since, 3125° lies in second quadrant, where the value of cos 315° is –ve
2
1° –1
\ cos 157 2 = 2 ­2 + 2

4. If sin A = 1 p + 1 show that sinA = –1 p3 + 1
Solution : 3 2 p 2 p3

Here, sin A = 1 p + 1
3 2 p

Now, sinA = 3sin A – 4sin3 A
3 3
3
1 1 1 1
= 3 × 2 p+ p ­– 4 2 p + p

=23 p + 1 – 4 × 1 p3 + 3p . 1 p+ 1 + 1
p 8 p p p3

=23 p + 1 – 1 P3 – 3 p+ 1 – 1 × 1 = –1 p3 – 1 1
p 2 2 p 2 p3 2 2 p3

= ­– 1 p3 + 1 = R.H.S. proved.
2 p3

204 Infinity Optional Mathematics Book - 10

5. Prove the following:
Solution:

(a) tan 45° + A = 1 + sinA
2 1 – sinA

L.H.S. = tan 45° + A
2

= sin 45° + A
cos 45° + 2
A
2 AA

= sin45° . cos 2 + cos45° . sin 2
cos A A
cos45° . 2 – sin45° . sin 2

= 1 cos A + 1 . sin A
2 2 2 2
1 1
2 cos A – 2 . sin A
2 2

cos A + sin A
2 – 2
= A A
cos 2 sin 2

cos A + sin A 2
2 2
= A A 2
2 2
cos – sin

= cos A + 2sin A . cos A + sin2 A
2 2 2 2
A A A A
cos2 2 – 2cos 2 . sin 2 + sin2 2

= 1 + sinA
1 – sinA

= R.H.S. Proved

( b) 1 +sinco2sA2A . 1 cosA = tan A
+ cosA 2

L.H.S. = 1 sin2A . 1 cosA
+ cos2A + cosA

= 2sinA . cosA . 1 cosA
2cos2A + cosA

2sin A . cos A
2 2
= A
2cos2 2

Infinity Optional Mathematics Book - 10 205

= sin A = tan A = R.H.S. proved
cos 2 2
A
2

(c) cot A + 45° – tan A = 2cosA
2 2 – 45° 1 + sinA

L.H.S. = cot A – tan A – 45° = 2cosA
2 + 45° 2 1 + sinA
A A
= cot 2 . cot45° – 1 – tan 2 – tan45°

cot A + cot45° 1 + tan A . tan45°
2 2

= cot A – 1 – tan A – 1
2 + 1 2
A A
cot 2 1 + tan 2

= 1 – tan A – tan A – 1
1 + 2 2
A A
tan 2 1 + tan 2

= 1 – tan A – tan A + 1
= 2 2
A
1 + tan 2
–
2 1 tan A
2
A
1 + tan 2

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

= 2 cos A – A cos A + A
2 sin 2 2 sin 2
A A × A A
cos 2 sin 2 cos 2 sin 2
+ +

= 2 cos2 A – sin2 A
2 2
A A A A
cos2 2 + 2cos 2 . sin 2 + sin2 2

= 2cosA
1 + sinA

= R.H.S. proved

206 Infinity Optional Mathematics Book - 10

(d) 12 cot θ – θ = cot q
2 tan 2

L.H.S = 1 cot θ – tan θ
2 2 2

1 cos θ sin θ
2 2 2
= sin θ – θ
cos 2
2

1 cos2 θ – sin2 θ
2 2 2
= sin θ θ
2 .cos 2

= cos q cosθ = cotq = R.H.S. proved
= sinθ
2sin θ . cos θ
2 2

Exercise 5.2

1. Find the value of sinA, cosA and tanA when

(a) sinA2 = 3 (b) cosA2 = 153 (c) tanA2 = 4
5 3

2. Find the value of sinA, cosA and tanA when

(a) sinA3 = 12 (b) cosA3 = 1 (c) tanA3 = 1
2

3. (a) If cosA2 = 12, show that sinA = 3 . (b) If tan θ = 1 , show that tanq = 3
2 2 3

(c) If sin A = 53, show that sinA = 111275 (d) If cos θ = 2 , show that cosq = – 22
3 3 3 27

(e) If tan q = 15, prove that tanq = 37
3 55

4. (a) If cos30º = 3 , show that
2

(i) sin15º = 3 –1 (ii) cos15º = 3 + 1 (iii) tan15º = 2 – 3
22 2 2

(b) If cos45º = 1 , show that
2

(i) sin22 1° = 1 2– 2 (ii) cos2212° = 1 2 + 2 (iii) tan2212° = 3–2 2
2 2 2

(c) If cos330º = 3 , prove that
2

Infinity Optional Mathematics Book - 10 207

(i) sin165º = 3 –1 (ii) cos165º = –   3 +1
22 22

5. (a) If cos θ = 1 p + 1 , show that cosq = 1 p2 + 1
2 2 p 2 p2

(b) If cos θ = 1 a+ 1 , show that cosq = 1 a3 + 1
3 2 a 2 a3

(c) If sin θ = 1 b+ 1 , show that cosq = – 1 b2 + 1
2 2 b 2 b2

(d) If sin θ = 1 x+ 1 , show that sinq = – 1 x3 + 1
3 2 x 2 x3

6. Prove the following identities:

(a) 1 +sincoAsA = cotA2 (b) 1 –sincoAsA = tanA2

(c) 1 +sincoθsθ = tan θ (d) 1 –sincos = cot 2
2

(e) 11 –+ cosA = cot2A2 (f) 11 +– ccooss = tan2 2
cosA

(g) cosecβ + cotβ = cot β (h) 2sisnin2θθ = cos2 2θ – sin2 θ
2 2

(i) cos3A2 – sin3A2 =1+ 1 sinA (j) 2sinA + sin2A = cot2A2
cosA2 – sinA2 2 2sinA – sin2A

(k) cotA2 – tanA2 = 2cotA (l) cot θ + tan θ
2 2
= secθ
θ θ
cot 2 – tan 2

(m) cosA2 – sinA2 = secA – tanA (n) 1 – sinθ = cos θ + sin θ
cosA2 + sinA2 2 2

(o) sesceAcA+ 1 = 2cos2A2

(p) sin θ . cos3 θ + cos θ . sin3 θ = 1 sinθ (q) 1 – 2sin2  – A = sinA
2 2 2 2 2 4 2

2tan  – θ = cosq (s) 1 – tan2  – θ = sin θ
(r) 4 2 4 4 2

1 +tan2 4 – θ 1 + tan2  – θ
2 4 4

208 Infinity Optional Mathematics Book - 10

(t) cos2  – θ – sin2  – θ = sin2θ
4 4 4 4

7. Prove the followings:

(a) 11 +– cosθ + sinθ = cot θ (b) 1 – cosθ + sinθ = tan θ
cosθ + sinθ 2 1 + cosθ + sinθ 2

θ
sinθ + sin2
(c) + cosθ + cos θ = tan θ (d) 1 sin2θ × 1 cosθ = tan θ
2 2 + cos2θ + cosθ 2
1

(e) tan 45° + A = secA + tanA = 1 + sinA
2 1 – sinA

(f) tan 45° – A = secA – tanA = 1 – sinA
2 1 + sinA

θθ
(g) sec 45° + 2 × sec 45° – 2 = 2secq

(h) cot θ + 45° – tan θ – 45° = 2(secq – tanq)
2 2

(i) tan c + θ + tan c – θ = 2secq
4 2 4 2

θ – 1 + sinθ
(j) csions22θ – 1 + sinθ θ
= cot2

(k) cos6  + sin6  = 1 – 3 sin2α
2 2 4

(l) (cosα + cosβ)2 + (sinα + sinβ)2 = 4cos2 – β
2

(m) (cosα – cosβ)2 + (sinα – sinβ)2 = 4sin2  – β
2

Infinity Optional Mathematics Book - 10 209

5.3 Transformation of Trigonometric Formulae

The sum or difference form of sine and cosine ratios can be transformed into

the product form and the product form can also be transformed into the sum or

difference form.

Transformation of products into sum or difference
We know that
sin (A + B) = sinA . cosB + cosA . sinB … (i)
sin (A – B) = sinA . cosB – cosA . sinB … (ii)
cos (A + B) = cosA . cosB – sinA . sinB … (iii)
cos (A – B) = cosA . cosB + sinA . sinB … (iv)

Now adding (i) and (ii) we get
Sin (A + B) + sin (A – B) = 2sinA . cosB

\ 2sinA . cosB = sin(A + B) + sin(A – B)

Subtracting (ii) from (i), we get
Sin (A + B) – sin (A – B) = 2cosA . sinB

\ 2cosA . sinB = sin(A + B) – sin(A – B)

Adding (iii) and (iv), we get
Cos(A + B) + cos (A – B) = 2cosA . cosB

\ 2cosA . cosB = cos(A + B) + cos(A – B)

Subtracting (iii) from (iv), we get
cos(A – B) – cos (A + B) = 2sinA . sinB
\ 2sinA . sinB = cos(A – B) – cos(A + B)

Transformation of sum or difference into product.

From the above formula, we have
sin (A + B) + sin (A – B) = 2sinA . cosB … (i)
sin (A + B) – sin (A – B) = 2cosA . sinB … (ii)
cos (A + B) + cos (A – B) = 2cosA . cosB … (iii)
cos (A – B) – cos (A + B) = 2sinA . sinB … (iv)

Let A + B = C and A ­ – B = D

then A = C + D and B = C – D
2 2

Now, substituting the values of A, B, A + B and A – B in (i), (ii), (iii) and (iv), we get

sinC + sinD = 2 sinC + D . cos C – D
2 2

210 Infinity Optional Mathematics Book - 10

sinC – sinD = 2 cos C + D . sin C – D
cosC + cosD 2 2
cosD – cosC
or, cosC – cosD = 2 cos C + D . cos C – D
2 2

= 2 sin C + D . sin C – D
2 2

= – 2 sin C + D . sin C – D
2 2

Transformation formulae in Tabular form
Transformation of product into sum or difference.

2sinA.cosB = sin(A + B) + sin (A – B)
2cosA.sinB = sin(A + B) – sin (A – B)
2cosA.cosB = cos(A + B) + cos (A – B)
2sinA.sinB = cos(A – B) – cos (A + B)

Transformation of sum or difference into product

sinC + sinD = 2 sin C + D . cos C – D
2 2

sinC – sinD = 2 cos C + D . sin C – D
2 2

cosC + cosD = 2 cos C + D . cos C – D
2 2

cosC – cosD = – 2 sin C + D . sin C – D
2 2

WORKED OUT EXAMPLES

1. Express the following products into sum or difference.

(a) sin28° . cos16° (b) sin5q . cos7q (c) 2cos40°. Cos80°

Solution:

(a) sin28º . cos16º (b) sin5q . cos7q

= 21(2sin28º.cos16º) = 21(2cos7q . sin5q)

= 21[sin (28º + 16º) +sin (28º – 16º)] = 21[ sin (7q + 5q) – sin (7q – 5q)]
= 21(sin44º + sin12º) = 21(sin12q – sin2q)
(c) 2cos40º . cos80º

= 2cos80º . Cos40º

= cos (80º + 40º) + cos (80º – 40º)

= cos120º + cos40º

= cos40º – 1
2

Infinity Optional Mathematics Book - 10 211

2. Express the followings into product form

(a) sin50° + sin40° (b) cos140° + cos50°

(c) sin70° – cos80° (d) cos50° – cos70°

Solution:

(a) Sin50º + sin40º

= 2 sin 50° + 40° . cos 50° – 40°
2 2

= 2sin45º . cos5º

= 2 × 1 cos5º
2

= 2 cos5º



(b) Cos140º + cos50º

= 2cos140°2+ 50° . cos140°2– 50°
= 2cos95º . cos45º

= 2cos95º . 1
2

= 2 cos95º

(c) Sin70º – cos80º

= sin70º – cos (90 – 10º)

= sin70º – sin10º

= 2cos 70° + 10° . sin 70° – 10°
2 2
= 2cos40º . sin30º

= 2cos40º. 1
2
= cos40º

(d) Cos50º – cos70º

= – (cos70º – cos50º)

=– – 2 sin70° + 50° . sin70° – 50°
2 2

= 2sin60º. Sin10º

= 2 × 3 sin10º
2

= 3 sin10º

3. Prove the following :

(a) Sin100º – sin40º – sin20º = 0
(b) 2sin (45° + q) . cos (45 – q) = 1 + sin2q

212 Infinity Optional Mathematics Book - 10

(c) csoins((8800°° + A) – sin(80° – A) = tanA.
+ A) + cos(80° – A)

Solution:

(a) L.H.S. = sin100º – sin40º – sin20º

= 2 cos100°2+ 40° . sin100°2– 40° – sin20º
= 2cos70º. Sin30º – sin20º

= 2 × 1 cos (90º – 20º) – sin20º
2

= sin20º – sin20º

= 0 = R.H.S. proved

(b) L.H.S. = 2 sin (45° + q). cos (45° –q)

= sin (45° + q + 45° – q) + sin (45° + q – 45° + q)

= sin90º + sin2q

= 1 + sin2q = R.H.S. proved

(c) L.H.S. = sin(80° + A) – sin(80° – A)
cos(80° + A) + cos(80° – A)

2cos 80° + A + 80° – A . sin 80° + A – 80° + A cos80° × sinA
2 2 cos80° × cosA
= =
2cos 80° + A + 80° – A 80° + A – 80° + A
2 . cos 2

= sinA = tanA = R.H.S. proved
cosA

4. Prove the following :

(a) sinq + sin3q + sin5q + sin7q = 4cosq . cos2q . sin4q

(b) sin92θ . sin32θ + sin52θ . sin θ = sin4q .sin2q
2

(c) csoinsAA – cos2A + scoins33AA = cot2A
– sin2A +

(d) scions42AA . cos3A – cos2A . cos7A = sin7A + sin3A
. sin3A – sin2A . sin 5A sinA

(e) (sinA – sinB)2 + (cosA + cosB)2 = 4cos2 A+B
2

Infinity Optional Mathematics Book - 10 213

(f) 8cos40º . cos100º . cos160º = 1

(g) sin10º . sin30º . sin50º . sin70º = 1
16

(h) sinA sin2A – sin2B cosB = tan (A + B)
. cosA – sinB .

(i) sin2q. cos3q = 1 (2cosq – cos3q – cos5q)
16

Solution:

(a) L.H.S. = sinq + sin3q + sin5q + sin7q

= (sin7q + sinq) + (sin 5q + sin3q)

=2 sin 7θ + θ . cos 7θ – θ + 2sin 5θ + 3θ . cos 5θ – 3θ
2 2 2 2

= 2sin4q . cos3q + 2sin4q.cosq

= 2sin4q (cos3q + cosq)

= 2sin4q.2cos 3θ + θ . cos 3θ – θ
2 2

= 4cosq.cos2q . sin4q = R.H.S. proved

(b) L.H.S. = sin92θ.sin32θ + sin52θ . sin2θ

= 1 2sin92θ.sin32θ + 2sin52θ . sinθ2
2

= 1 cos 9θ – 3θ – cos 9θ + 3θ + cos 5θ – θ – cos 5θ + θ
2 2 2 2 2 2 2 2 2

= 12(cos3q – cos6q + cos2q – cos3q)

= –1 (cos6q – cos2q)
2

= –1 × – 2sin6θ + 2θ . sin 6θ – 2θ
2 2 2

= sin4q.sin2q = R.H.S. proved

(c) L.H.S. = cosA – cos2A + cos3A
sinA – sin2A + sin3A

= cos3A + cosA – cos2A
sin3A + sinA – sin2A

214 Infinity Optional Mathematics Book - 10

= 2cos3A2+ A . cos3A2– A – cos2A
2sin3A2+ A . cos3A2– A –

sin2A

= 2cos2A . cosA – cos2A
2sin2A. cosA – sin2A

= cos2A(2cosA – 1)
sin2A(2cosA – 1)

= cot2A = R.H.S. proved

(d) L.H.S. = cos2A . cos3A – cos2A . cos7A
sin4A . sin3A – sin2A . sin5A

= 2cos3A . cos2A – 2cos7A . cos2A
2sin4A . sin3A – 2sin5A . sin2A

= cos(3A + 2A) + cos(3A – 2A) – cos (7A + 2A) – cos(7A – 2A)
cos(4A – 2A) – cos(4A + 3A) – cos(5A – 2A) + cos(5A + 2A)

= cos5A + cosA – cos9A – cos5A
cosA – cos7A – cos3A + cos7A

= cosA – cos9A
cosA – cos3A

= cos9A – cosA
cos3A – cosA

= – 2sin9A2+ A . sin9A2– A
– 2sin3A2+ A . sin3A2– A

= sin5A . sin4A
sin2A . sinA

= sin5A . 2sin2A . cos2A
sin2A . sinA

= 2sin5A . cos2A
sinA

= sin(5A + 2A) + sin(5A – 2A)
sinA

= sin7A + sin3A
sinA

= R.H.S. proved

Infinity Optional Mathematics Book - 10 215

(e) L.H.S. = (sinA – sinB)2 + (cosA + cosB)2

= 2cosA + B . sinA – B 2 2cosA + B . cosA – B 2
2 2 2 2
+

= 4cos2A + B . sin2A – B + 4cos2A + B . cos2A – B
2 2 2 2

= 4cos2A + B sin2 A – B + cos2A – B
2 2 2
4cos2A + B
= 2

= R.H.S. proved

(f) L.H.S. = 8cos40º . cos100º . cos160º

= 4cos40º (2cos160º . cos100º)

= 4cos40º [cos (160° + 100°) + cos (160° – 100°)]

= 4cos40º (cos260º + cos60º)

= 4cos260º . cos40º + 4 × 1 cos40º
2

= 2(2cos260º. Cos40º) + 2cos40º

= 2[cos(260 + 40º) + cos (260–40º)] + 2cos40º

= 2(cos300º + cos220°) + 2cos40º

= 2cos (90° × 3 + 30°) + 2cos (90° × 2 + 40°) + 2cos40º

= 2sin30º – 2cos40º + 2cos40º = 2 × 1
2

=1

= R.H.S. proved

Alternate method

L.H.S. = 8cos40º . cos100º . cos160º

= 8cos40º . cos(60º + 40°) . cos(120° + 40º)

= 8cos40º.(cos60° . cos40°–sin60°. sin40°)(cos120°. cos140°– sin120° . sin40°)

= 8cos40º 21cos40° – 3 sin 40° – 21 cos40° – 3 sin 40°
2 2

= – 8cos40º 41cos240° – 3 sin240°
4

= – 48cos40°(cos240° – 3 + 3cos240°)

= – 2(4cos340° – 3cos40°)

216 Infinity Optional Mathematics Book - 10

= – 2 × cos120° [∴ 4cos3 – 3coA = cos3A]

= – 2 × – 1 = 1
2

= R.H.S. proved

(g) L.H.S. = sin10º . sin30º . sin50º . sin70º

= 1 . 1 sin10º (2sin70º . sin50º)
2 2
1
= 4 sin10º [cos (70° – 50°) – cos (70° + 50°)]

= 1 sin10º (cos20º – cos120º)
4
1 1
= 4 sin10º . cos20º + 8 sin10º

= 1 × 1 (2cos20º . sin10º) + 1 sin10º
4 2 8
1 1
= 8 [sin (20° + 10º) – sin(20° – 10°)] + 8 sin10º

= 1 sin 30º – 1 sin 10º + 1 sin 10º
8 8 8
1 1
= 8 × 2

= 1 = R.H.S. Proved
16

Alternate method

L.H.S. = sin10º . sin30º . sin50º . sin70º

= 1 sin10º . sin(60º – 10º) . sin(60° + 10°)
2
1
= 2 sin10º(sin60° . cos10° – cos60° . sin10°)(sin60° . cos10° + cos60° . sin10°)

= 1 sin10º 3 cos10° – 21 sin10° 3 cos10° + 21 sin10°
2 2 2

= 1 sin10º 3 cos210° – 41 sin210°
2 4
1
= 8 × sin10°(3 – 3sin210º – sin210º)

= 1 (3sin10º – 4sin310º)
8
1
= 8 sin30° [∴ 3sinA – 4sin3A = sin 3A]

= 1 × 1 = 1 = R.H.S. Proved
8 2 16
sin2A – sin2B
(h) L.H.S. = sinA . cosA – sinB . cosB

= 2sin2A – 2sin2B cosB
2sinA . cosA – 2sinB .

= 1 – cos2A – 1 + cos2B
sin2A – sin2B

Infinity Optional Mathematics Book - 10 217

= – (cos2A – cos2B)
sin2A – sin2B

=– – 2sin2A + 2B . sin2A – 2B = sin(A + B) . sin(A – B)
2 2 cos(A + B) . sin(A – B)

2cos2A + 2B . sin2A – 2B
2 2

= tan (A + B) Proved

(i) R.H.S. = 1 (2cosq – cos3q – cos5q)
16

= 1 {2cosq – (cos5q + cos3q}
16

= 1 2cosθ – 2cos5θ 2+ 3θ . sin5θ – 3θ
16 2

= 1 (2cosθ – 2cos4θ . cosθ)
16

= 1 × 2cosq (1 – cos4q)
16

= 1 cosq (2sin22q)
8

= 41cosq (2sinq . cosq)2
= 41cosq 4sin2q . cos2q

= sin2q . cos3q

= L.H.S. proved

5. If sin2A + sin2B = 1 and cos2A + cos2B = 1 show that tan (A + B) = 2
3 2 3
Solution :

Here, sin2A + sin2B = 1
3

or, 2sin2A + 2B . cos 2A – 2B = 1
2 2 3

or, 2 sin (A + B) . cos (A – B) = 1 …………….. (i)
3

And, cos2A + cos2B = 1
2

or, 2cos 2A + 2B . cos 2A – 2B = 1
2 2 2

or, 2cos (A + B) . cos(A – B) = 1 …………….. (ii)
2

Dividing (i) by (ii) we get

218 Infinity Optional Mathematics Book - 10

1

2sin(A + B) . cos (A – B) = 3
2cos(A + B) . cos (A – B) 1

2 2
3
\ tan (A + B) = proved.

6. Prove that : 2cos 13. cos193 + cos133 + cos153 = 0
Solution :

L.H.S. = 2cos13. cos193 + cos133 + cos153

= cos 9 +  + cos 9 –  + cos133 + cos153
13 13 13 13

= cos1103 + cos183 + cos133 + cos153

= cos  – 3 + cos  – 5 + cos133 + cos153
13 13

= – cos133 – cos153 + cos133 + cos 5
13

= 0 = R.H.S. Proved.

Alternative Method:
= 2cos13. cos193 + cos133 + cos153

= cos 9 +  + cos 9 –  + cos133 + cos153
13 13 13 13

= cos1103 + cos183 + cos133 + cos153

= cos 10 + cos133 + cos183 + cos 5
13 13

10 + 3 10 – 3 8 + 5 8 – 5
13 13 13 13 13 13 13 13
= 2cos . cos + 2cos . cos
2 2 2 2

= 2cos1236 . cos276 + 2cos 13 . cos236
26

= 2cos 13 cos276 + cos 3
26 26

= 2cos2 cos276 + cos 3
26

= 2 × 0 cos276 + cos236
= 0 = R.H.S. proved

Infinity Optional Mathematics Book - 10 219

Exercise 5.3

1. Express each of the following of a sum or difference :

(a) 2sin25º. Cos5º (b) 2cos65º. Sin25º (c) cos35º. Cos25º
(f) 2 sin7q.cos9q
(d) sin100º . sin 80º (e) 2sin 5q . cos3q

(g) 2sin5q . cos7q (h) sin9A.sin12A

2. Express each of the following as a product :

(a) sin70º + sin20º (b) sin55º – sin25º (c) cos9q + cos3q
(d) cos11q – cos7q (e) sin9q – sin11q (f) cos110º – cos40º
(g) cos40º + cos140º (h) sin35º + sin115º
3. Prove the followings :

(a) 2sin8A.cos3A = sin11A + sin5A

(b) sin3q. Cos9q = 1 (sin12q – sin6q)
2

(c) cos55º . cos65º = 1 cos10° – 1
2 2

(d) sin80º . sin20º = 1 1 – cos100°
2 2

(e) 2sin (45° + q) . cos (45° – q) = 1 + sin2q.

4. Prove the followings :

(a) sin70º + sin20º = 2 cos25º

(b) sin20º – sin50º = – 2cos35° sin15º

(c) cos8q – cos10q = 2 sin9q.sinq (d) cos100 + cos20º = cos40º

(e) sin105º + sin15º = 3 (f) sin75º – sin15º = 1
2 2

(g) cos105º + sin75º = 1
2

(h) sin(150° + A) + sin (150° – A) = cosA

(i) cos (45° + q) + cos (45° – q) = 2 cos q (j) cos20º + cos100º + cos140º = 0

(k) cos70º + cos50º – cos10º = 0 (l) sin70º + cos100º – cos40º = 0

220 Infinity Optional Mathematics Book - 10

5. Prove the followings :

(a) csoins + sinβ = tan + β (b) csoins55θθ ++ csoins33θθ = tan4q
+ cosβ 2

(c) csoins7755°° + cos15° = 3 (d) ccooss8800°° +– scoins1100°° = tan35º
– sin15°

(e) csoins((4455°° + A) + cos(45° – A) = 1 (f) sin2A + sin2B = tan(A + B)
+ A) + sin(45° – A) sin2A – sin2B tan(A – B)

(g) csoins((8800°° + A) – sin(80° – A) = tanA
+ A) + cos(80° – A)

(h) ccoossAA – cosB = – tan A + B . tan A – B
+ cosB 2 2

(i) ssiinnθθ + sinβ = tan θ +β . cot θ –β
– sinβ 2 2

6. Prove the followings :

(a) sin7A . sin3A + sin11A . sinA = sin8A . sin4A
(b) sin3A . cosA + cos4A . sin2A = sin5A . cosA

(c) cos39º . sin18º + cos15º . sin6º = sin24º . cos33º

(d) sin92A . sin32A + sin 5A . sinA2 = sin4A . sin2A
2

(e) sin7A + sin5A + sin3A + sinA = 4cosA . cos2A . sin4A

(f) cosA + cos3A + cos5A + cos7A = 4cosA . cos2A . cos4A

(g) cos30º + cos50º + cos70º + cos90º = 2cos10.cos20º

(h) cos30º – cos50º + cos70º – cos90º = 2 3 sin10º . cos20º

(i) sin50º + sin40º + sin20º + sin10º = sin70º + sin80º
(j) 2cos13 . cos193 + cos133 + cos153 = 0
7. Prove the following identities :

(a) (sin9θ + sin3θ) + (sin7θ + sin5θ) = tan6q
(cos9θ + cos3θ) + (cos7θ + cos5θ)

(b) sin2θ + sin3θ + sin5θ + sin6θ = tan4q
cos2θ + cos3θ + cos5θ + cos6θ

(c) csoinsAA + sin2A + sin4A + sin5A = tan3A
+ cos2A + cos4A + cos5A

(d) ((ssiinn47θθ++ssiinn25θθ))((ccooss4θθ––ccooss58θθ)) = 1 (e) (cosθ – cos3θ) (sin8θ + sin2θ) = 1
(cos4θ – cos6θ) (sin5θ – sinθ)

Infinity Optional Mathematics Book - 10 221

(f) csoins55AA + sin2A – sinA = tan2A (g) cosA – cos2A + cos3A = cot2A
+ cos2A + cosA sinA – sin2A + sin3A

(h) scoins44AA + sin3A + sin2A = tan3A (i) sin8θ . cosθ – sin6θ . cos3θ = tan2 q
+ cos3A + cos2A cos2θ . cosθ – sin4θ . sin3θ

(j) csoins22θθ . sinθ + sin6θ . cos3θ = tan5 q (k) ccooss44θθ . cos3θ – cos5θ . cos2θ = tan2 q
. sinθ + cos6θ . sin3θ . sin3θ – cos5θ . sin2θ

(i) cos2θ . cos3θ – cos2θ . cos7θ = sin7θ + sin3θ
sin4θ . sin3θ – sin2θ . sin5θ sinθ

8. Prove the following:

(a) sin20º . sin40º . sin80º = 83 (b) 8cos20º . cos40º . cos80º = 1

(c) cos10º . cos50º . cos70º = 83 (d) 8sin10º . sin50º. sin70º = 1

(e) sin20º . sin30º . sin40º . sin80º = 136

(f) 16cos10º . cos30º . cos110°. cos130º = 3

(g) 16cos40º . cos60º . cos100º . cos160º = 1 (h) Cos12º . cos24º. cos48° . cos84º = 1
16

(i) Sin6º . sin42º . sin66º . sin78º = 116 (j) tan20º . tan40º . tan80º = 3

9. Prove the following:

(a) (cosA + cosB)2 + (sinA + sinB)2 = 4cos2 A–B
2

(b) (cos2B – cos2A)2 + (sin2A – sin2B)2= 4sin2(A – B)

(c) sin2  + θ – sin2  – θ = 1 sin q
8 2 8 2 2

(d) cos2A + cos2 (A + 120º) + cos2 (A – 120º) = 3
2

(e) sin2A + sin2 (A + 120º) + sin2 (A – 120º) = 3
2
cos2θ sin2
(f) sin . cos – sinθ . cosθ = cot (α + q)
+

(g) sinA sin2A – sin2B cosB = tan (A + B)
. cosA – sinB .

(h) cos3 q.sin2 q = 116(2cos q – cos3 q – cos5 q)

(i) sin4 q.cos2 q = 312(cos6 q – 2cos4 q – cos2 q + 2)
(j) cos215c . cos415c . cos815c . cos1145c =
1
16

222 Infinity Optional Mathematics Book - 10

(k) 1 + cos c 1 + cos38c 1 + cos58c 1 + cos78c = 1
8 8

(l) 16 cos3q = 16 cos5q + 2 cosq – cos3q – cos5q

10. (a) If cosA + cosB = 1 and sinA + sinB = 41 , prove that
2

(i) tan A+B = 1 (ii) cos (A + B) = 53.
2 2

(b) If sinA + sinB = a and cosA + cosB = b, then prove that cot A+B = b
2 a

5.4 Conditional Trigonometric Identities

Let us consider a trigonometric relation sec²q – tan²q = 1
It is true for any value of q. So, it is said to be trigonometric identity. An identity
which is true on the particular condition is called a conditional identity. The
condition given in this section is the sum of three angles in a triangle is 180º or pc.

WORKED OUT EXAMPLES

1. If A + B + C = πc then show that
(a) cotA . cotB + cotB . cotC + cotA . cotC = 1
(b) tan2A + tan2B + tan2C = tan2A . tan2B . tan2C

Solution:
(a) Here, A + B + C = pc
or, A + B = pc – C

Now, cot (A + B) = cot (pc – C) = – cotC
or, coctoAtA. c+otcBotB– 1 = – cotC
or, cotA . cotB – 1 = –cotC . cotA – cotB . cotC

\ cotA . cotB + cotB . cotC + cotA . cotC = 1 proved

(b) Here, A + B + C = pc

A + B = pc – C

or, 2A + 2B = 2pc – 2C [multiplying both sides by 2]

Now, tan (2A + 2B) = tan (2pc – 2C) = –­ tan2C

Infinity Optional Mathematics Book - 10 223

or, 1 tan2A + tan2B = – tan2C
– tan2A . tan2B

or, tan2A + tan2B = – tan2C + tan2A . tan2B . tan2C

\ tan2A + tan2B + tan2C = tan2A . tan2B . tan2C Proved.
2. If A + B + C = pc, prove that Sin2A + sin2B + sin2C = 4sinA. sinB. sinC

Solution :

Here, A + B + C = pc

A + B = pc – C

Now, sin (A + B) = sin (pc – C) = sinC

And cos (A + B) = cos (pc – C) = – cosC

L.H.S. = sin2A + sin2B + sin2C

= 2sin 2A + 2B . cos 2A – 2B + sin2C
2 2

= 2sin(A + B) . cos (A – B) + sin2C

= 2sinC . cos (A – B) + 2 sinC . cosC

= 2sinC {cos(A – B) + cosC}

= 2sinC {cos (A – B) – cos (A + B)}

= 2sinC × 2sinA . sinB

= 4sinA . sinB . sinC

= R.H.S. proved.

3. If A + B + C = 180º, prove that cos2A – cos2B – cos2C = – 1 + 4cosA. sinB. sinC

Solution :

Here, A + B + C = 180º

or, A + B = 180º – C

Now, sin (A + B) = sin (180º – C) = sinC

And cos (A + B) = cos (180º – C) = – cosC

L.H.S. = cos2A – cos2B – cos2C

= – 2sin 2A + 2B . sin 2A – 2B – cos2C
2 2

= – 2sin(A + B) . sin (A – B) – cos2C
= – 2sinC . sin (A – B) – 1 + 2 sin2C

= –1 – 2sinC {sin(A – B) – sinC}

= – 1 – 2sinC {sin (A – B) – sin (A + B)}

224 Infinity Optional Mathematics Book - 10

= – 1 + 2sinC (sin(A + B) –­ sin (A – B) }
= –1 + 2sinC × 2 cosA . sinB
= –1 + 4cosA . sinB . sinC
= R.H.S. proved.

4. If A, B and C are the vertices of ∆ABC, then prove that

cos2A + cos2B – sin2C = – 2cosA . cosB. cosC

Solution:

Here, A + B + C = 180º

or, A + B = 180º – C

Now, sin (A + B) = sin (180º – C) = sinC

And cos (A + B) = cos (180º – C) = – cosC

L.H.S. = cos2A + cos2B – sin2C

= 1 + cos2A + 1 + cos2B – sin2C
2 2

= 1 + 1 cos2A + 1 + 1 cos2B – sin2C
2 2 2 2

= 1 + 12(cos2A + cos2B) – sin2C

= 1 + 1 × 2 cos 2A + 2B . cos 2A – 2B – sin2C
2 2 2

= 1 + cos (A + B) . cos(A – B) – sin2C

= 1 – cosC . cos(A – B) – 1 + cos2C

= – cosC {cos(A–B) – cosC}

= – cosC {cos (A–B) + cos (A + B)}

= – cosC × 2cosA . cosB

= – 2cosA . cosB . cosC

= R.H.S. proved

Alternate method

Here, A + B + C = 180º

A + B = 180º – C

Now, cos(A + B) = cos(180º – C)

Infinity Optional Mathematics Book - 10 225

or, cosA . cosB – sinA . sinB = – cosC
or, cosA . cosB + cosC = sinA . sinB
Now squaring on both sides, we get
or, (cosA . cosB + cosC)2 = (sinA . sinB)2
or, cos2A . cos2B + 2cosA . cosB . cosC + cos2C = (1 – cos2A) (1– cos2B)
or, cos2A . cos2B + 2cosA . cosB . cosC + 1 – sin2C

= 1 – cos2A – cos2B + cos2A . cos2B
∴ cos2A + cos2B – sin2C = –2cos A cosB cos C. Proved.

5. If A + B + C = pc, prove that

– SinA + sinB + sinC = 4cosA . sinB . sinC
222

Solution : Here, A + B + C = pc

A + B = pc – C

or, A + B = c – C [Dividing both sides by 2)
2 2 2 2

Now, sin A + B = sin c – C = cos C
2 2 2 2 2

cos A + B = cos c – C = sin C
L.H.S. 2 2 2 2 2

= – sinA + sinB + sinC

= ­ – (sinA – sinB) + sinC

= – 2cosA + B . sin A – B + sinC
2 2

= – 2cos A + B . sin A – B + sinC
2 2 2 2

= – 2sin C . sin A – B + 2sin C . cos C
2 2 2 2 2

= – 2sinC2 sin A – B – cos C
2 2 2

= ­–2sinC2 sin A – B – sin A + B
2 2 2 2

= 2sinC2 sin A + B – sin A – B
2 2 2 2

= 2sin C × 2cosA2 . sinB2
2

= 4cosA2 . sin B . sinC2 = R.H.S. proved
2

226 Infinity Optional Mathematics Book - 10

6. If A + B + C = 180º, prove that sin2 A – sin2B2 + sin2C2 = 1 – 2 cosA2 × sinB2 × cosC2
2
Solution :

Here, A + B + C = 180º

Or, A + B = 180º – C

Or, A2 + B2 = 90° – C [Dividing both sides by 2)
2

Now, sin A + B = sin 90° – C = cosC2
2 2 2

cos A + B = cos 90° – C = sinC2
2 2 2

L.H.S. = sin2A2 – sin2 B + sin2C2
2

= 1 – cosA – 1 – cosB + sin2 C
2 2 2

= 1 – 1 cosA – 1 + 1 cosB + sin2 C
2 2 2 2 2

= – 1 (cosA – cosB) + sin2C2
2

=– 1 × – 2sin A + B . sin A – B + sin2C2
2 2 2

= sin A + B . sin A – B + sin2C2
2 2 2 2

= cosC2 . sin A – B + 1 – cos2C2
2 2

= 1 – cosC2 . cos C – sin A – B
2 2 2

= 1 – cosC2 . sin A + B – sin A – B
2 2 2 2

= 1 – cosC2 × 2cosA2 . sinB2
= 1 – 2cosA2 . sinB2 . cosC2

= R.H.S. proved

Alternate method

Here, A + B + C = 180º

or, A + C = 180º – B

We dividing both sides by 2, we get

A + C = 180° – B
2 2 2 2

Infinity Optional Mathematics Book - 10 227

Now, cos A + C = cos 90° – B
2 2 2

or, cos A . cos C – sin A . sin C = sin B
2 2 2 2 2

or, cos A . cos C – sinB2 = sin A . sin C
2 2 2 2

Now, squaring on both sides, we get

A C – sinB2 2 A . sinC2 2
2 2 2
cos . cos = sin

or, cos2A2 . cos2C2 – 2cos A . sinB2 . cosC2 + sin2 B = sin2 A . sin2 C
2 2 2 2

or, 1 – sin2 A 1 – sin2 C – 2cos A . sin B . cos C + sin2 B = sin2A2 . sin2 C
2 2 2 2 2 2 2

or, 1 – sin2A2 – sin2 C + sin2A2 . sin2 C – 2cos A . sinB2 . cosC2 + sin2B2 = sin2A2 . sin2C2
2 2 2

or, – sin2 A + sin2B2 – sin2 C = –1 + 2cos A . sinB2 . cosC2
2 2 2

∴ sin2 A – sin2B2 + sin2 C = 1 – 2cos A . sinB2 . cosC2 .
2 2 2
Proved

Note: For the question having square always take cos on both sides.

7. If A + B + C = 180º, prove that

sinA . cosB . cosC + cosA . sinB . cosC + cosA . cosB . sinC = sinA . sinB . sinC

Solution :

Here, A + B + C = 180º

or, A + B = 180º – C

Now, sin(A + B) = sin (180º – C)= sinC

Cos(A + B) = cos (180º – C) = – cosC

L.H.S. = sinA . cosB . cosC + cosA . sinB . cosC + cosA . cosB . sinC

= cosC (sinA . cosB + cosA . sinB) + cosA . cosB . sinC

= cosC . sin (A + B) + cosA . cosB . sinC

= cosC . sinC + cosA . cosB . sinC

= sinC (cosC + cosA . cosB)

228 Infinity Optional Mathematics Book - 10

= sinC {cosA . cosB – cos(A + B)}
= sinC (cosA . cosB – cosA . cosB + sinA . sinB)
= sinA . sinB . sinC
= R.H.S. proved

8. If A + B + C = pc, prove that

sinA2 + sinB2 + sin C = 1 + 4sinc – A . sinc – B . sinc – C
2 4 4 4

= 1 + 4 sin B + C . sinC + A . sinA + B
Solution: 444

R.H.S. = 1 + 4 sin B + C . sinC + A . sinA + B
4 4 4

= 1 + 4sinc – A . sinc – B . sinc – C
4 4 4
2sinc – A sinc – B sinc – C
=1+2 4 . 4 . 4

= 1 + 2 cosc –A – c + B – cosc – A + c – B . sinc – C
= 1 + A. 4 4 4
sinc
2cosB – – C – 2cos2 – A – B . sinc – C
4 4 4 4

= 1 + sin B – A + c – C – sin B – A – + C
4 4

– sin 2c – A –B + c – C – sin 2c – A –B – pc +C
4 4

= 1 + sinB + B – sin c – A– A – c – sin3c4– c – sin c – c + 2C
4 4 4
= 1 + sinB2 – –A  sinC2
sin 2 – sin 2 +

= 1 + sinB2 + sin A – 1 + sinC2
2
= sinA2 + sinB2 + sinC2 . L.H.S. proved

Alternate method

L.H.S. = sinA2 + sinB2 + sinC2 
sinB2 sinC2 2
= 1 + sinA2 + + – sin


= 1 + 2sin A+B . cos A–B + 2cos C+ . sin C–
4 4 4 4

= 1 + 2sin –C . cos A–B – 2cos +C . sin –C
4 4 4 4

= 1 + 2sin –C × cos A – B – cos  +C
4 4 4

Infinity Optional Mathematics Book - 10 229

= 1 + 2sin – C × – 2sin A– B+ + C . sin A– B– – C
4 8 8

= 1 – 4sin  – C . sin –B–B+ . sin A––+A
4 8 8

= 1 – 4sin  – C . sin  – B . sin A– 
4 4 4
 – A
= 1 + 4sin 4 . sin  – B . sin  – C = midle part proved
4 4

= 1 + 4sin B + C . sin C + A . sin A + B
4 4 4

= 1 + 4sin A + B . sin B + C . sin C + A = R.H.S.proved
4 4 4

Exercise 5.4

1. If A + B + C = c, prove that:

(a) tanA + tanB + tanC = tanA . tanB . tanC

(b) cotA . cotB + cotB . cotC + cotC . cotA = 1
(c) tanA2 . tanB2 + tanB2 . tanC2 + tanA2 . tanC2 = 1
(d) cotA2 + cotB2 + cotC2 = cotA2 . cotB2 . cotC2

(e) tan2A + tan2B + tan2C = tan2A . tan2B . tan2C

(f) cot2A . cot2B + cot2B . cot2C + cot2C . cot2A = 1

2. If A + B + C = c then prove that:
2

(a) tanA . tanB + tanB . tanC + tanC . tanA = 1

(b) cotA + cotB + cotC = cotA . cotB . cotC

3. If A + B + C = 180º, prove that :

(a) sinA + sinB + sinC = 4cosA2 . cosB2 . cosC2
(b) sinA – sinB + sinC = 4sinA2 . cosB2 . sinC2

(c) sinB – sinA + sinC = 4cosA2 . sinB2 . sinC2
(d) sinA – sinB – sinC = – 4cosA2 . sinB2 . sinC2
(e) cosA + cosB + cosC = 1 + 4sinA2 . sinB2 . sinC2
(f) cosB + cosC –­ cosA = 4sinA2 . cosB2 . cosC2 – 1

230 Infinity Optional Mathematics Book - 10

(g) cosA – cosB – cosC = 1 –­­ 4sinA2 . cosB2 . cosC2
4. If A + B + C = 180º, show that:
(a) sin2A + sin2B + sin2C = 4sinA . sinB . sinC
(b) sin2A – sin2B + sin2C = 4cosA . sinB . cosC
(c) cos2A + cos2B + cos2C = – 1 – 4 cosA . cosB . cosC
(d) cos2A + cos2B – cos2C = 1 – 4 sinA . sinB . cosC

5. If A, B and C are the three angles of a triangle ABC then prove that:
(a) sin2A2 + sin2B2 + sin2C2= 1 – 2 sinA2 . sinB2 . sinC2

(b) sin2A2 – sin2B2 + sin2C2 = 1 – 2 cosA2 . sinB2 . cosC2

(c) cos2A2 + cos2B2 + cos2C2 = 2 + 2 sinA2 . sinB2 . sinC2
(d) cos2A2 B cos2C2 2cosA2 cosB2 sinC2
+ cos2 2 – = . .

(e) cos2C2 + cos2B2 – cos2A2 = 2sinA2 . cosB2 . cosC2
6. If A + B + C = 180º, show that:

(a) sin2A + sin2B + sin2C = 2 + 2cosA . cosB . cosC

(b) sin2A – sin2B + sin2C = 2sinA . cosB . sinC

(c) cos2A + cos2B + cos2C = 1 – 2cosA . cosB . cosC

(d) cos2B + cos2C – cos2A = 1 – 2cosA . sinB . sinC

7. If A, B and C are the angles of a triangle, show that:

(a) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4sinA . sinB . sinC

(b) cos(B + C – A) + cos(C + A – B) + cos(A + B – C) = 1 + 4cosA . cosB . cosC

(c) sinBco.sAsinC + cosB + cosC = 2
sinC . sinA sinA . sinB

(d) cosA . sinB . sinC + cosB . sinC .sinA + cosC.sinA.sinB = 1+ cosA.cosB.cosC

(e) sinA.cosB.cosC + sinB.cosA.cosC + sinC.cosA.cosB = sinA.sinB.sinC

(f) sinA2 + sinB2 + sinC2 = 1 + 4sinc – A . sin c – B . sinc – C
4 4 4

(g) cosA2 + cosB2 + cosC2 = 4cosc – A . cos c – B . cosc – C
4 4 4

= 4cos B + C . cos C + A . cosA + B
4 4 4

(h) cos4A + cos4B + cos4C = –1 + 4cos2A . cos2B . cos2C

Infinity Optional Mathematics Book - 10 231

(i) ssinin2AA + sin2B + sin2C = 8sinA2 . sinB2 . sinC2
+ sinB + sinC

(j) cos22A + cos22B + cos22C = 1 + 2cos2A . cos2B . cos2C

8. If A + B + C = c then prove the followings :
2

(a) cos2A + cos2B + cos2C = 1 + 4sinA . sinB . sinC

(b) cos2A + cos2B ­ – cos2C + 1 = 4cosA . cosB . sinC

(c) sin2A + sin2B – sin2C = 4sinA . sinB . cosC

(d) sin2B + sin2C – sin2A = 4sinB . sinC . cosA

5.5 Trigonometric Equations

Let us consider the following relations
Sec2q – tan2q = 1 … (i)

sinq = 23 … (ii)

Let us consider q = 30º, 45º and 60º. Then putting these values of q in relation (i),

we get 1 2 1 2 4 1 3
3 3 3 3
Sec230º – tan230º = 3 – = – = = 1

Sec245º – tan245º = – ( 2 )2 – (1)2 = 2 – 1 = 1

Sec260º – tan260º = (2)2 – ( 3 )2 = 4 – 3 = 1

Here, relation (i) is true for every value of q. Such relations which satisfy every
value of the variable are called identities. So relation (i) is an identity.

Again putting the value of q in relation (ii), we get

Sin30º = 21, sin45º = 1, sin60º = 3
3 2
Here, relation (ii) is true only when q = 60º.

Such relations which satisfy some particular values of the variable are called
equations. Relation (ii) is an equation. Here relation (ii) involves the trigonometric
ration (ie. sinq), so, it is called a trigonometric equation.

Solution of trigonometric equations

We know that a trigonometric ratios of a certain angle has one and only one value.

But if the value of trigonometric ratio is given, the angle is not unique. To illustrate

the above fact, let us observe the graph of y = sinq and find the values of q when
1
y = sinq = 2

232 Infinity Optional Mathematics Book - 10

Y

0.1

0.8

0.6

0.4

0.2

X' O 30º 60º 90º 120º 150º 180º 210º 240º 270º 300º 330º 360º X
-0.2

-0.4

-0.6

-0.8

-1

Y' y = sinx

From the above graph, when sinq = 21, then the corresponding values of q are 30º, 150º
and so on.
180° – θ θ
Rules for finding angles :
SA
1. At first, we determine the quadrant where the angle falls. (360°+ θ)
For this we use the CAST table.

TC

(180° – θ) (360° – θ)

Quadrant I II III IV
Ratios
Sin and cosec +ve +ve –ve –ve
Cos ans sec +ve –ve –ve +ve
Tan and cot +ve –ve +ve –ve

2. Find the least positive angle of the trigonometric function in the first quadrant for the given
relation.

For example
(i) For sinq = 1 , the least positive angle in the first quadrant is 45º
2
(ii) For cosq = – 21, the least positive angle in the first quadrant for

Infinity Optional Mathematics Book - 10 233

cosq = 1 is 60º
2
For tanq = – 1 , the least positive angle in the first quadrant for tanq =
(iii) 3
(iv) 1 is 30º
3

For sinq = –1, the least positive angle in the first quadrant for sinq = 1

is 90º
3. When q is the least positive angle in the first quadrant then the angle in the

second quadrant = (180º – q)
The angle in the third quadrant = (180º + q)
The angle in the fourth quadrant = (360º – q)

To find the angle more them 360º we use (360º + q) and so on.

Note: To find the angle in the second, third and fourth quadrant, we add or
subtract the least positive angles of first quadrant in even multiple of 90º.

WORKED OUT EXAMPLES

1. Solve (0° ≤ q ≤ 360º) (b) 2cosq + 3 = 0 (c) 3 tanq – 3 = 0
(a) 3 sinq = 1

Solution

(a) Here, 2 sinq = 1
or, sinq = 1 (Here, sinq is positive. So, q lies in 1st and 2nd quadrants)

2

or, sinq = sin45º, sin(180º – 45º)

\ q = 45º, 135º

(b) Here, 2cosq + 3 = 0

or, cosq = –3 (Here cosq is negative. So q lies in 2nd and 3rd quadrant)
2

or, cosq = – cos30º

or, cosq = cos(180º – 30º), cos (180º + 30º)

\ q = 150º, 210º

(c) Here, 3 tanq – 3 = 0
or, tanq = 3 (Here, tanq is positive. So q lies in 1st and 3rd quadrants)

or, tanq = tan60º, tan (180º + 60º)

\ q = 60º, 240º

234 Infinity Optional Mathematics Book - 10

2. Solve (0° ≤ q ≤ 180º) (b) 9 tan2q – 3 = 0
(a) 4sin2q – 3 = 0

Solution:

(a) Here, 4sin2q – 3 = 0

or, sin2q = 3
4

or, sinq =± 3
2
Taking +ve sign,

sinq = 3
2

or, sinq = sin60º, sin (180 – 60º)

\ q = 60º, 120º

Taking –ve sign

sinq = – 2 3
or, sinq = – sin60°

or, sinq = sin(180º + 60º), sin (360º – 60º)

\ q = 240º, 300º

But 0º ≤ q ≤ 180º

Hence, q = 60º, 120º

(b) 9 tan2q – 3 = 0

or, tan2q = 1
or, tanq = ± 31

3

Taking +ve sign,

tanq = 1
3

or, tanq = tan30º, tan (180 + 30º)

\ q = 30º, 210º

Taking –ve sign

tanq = –1
3

or, tanq = – tan30º

or, tanq = tan (180º – 30º), tan (360º – 30º)

\ q = 150º, 330º

But, 0º ≤ q ≤ 180º

Hence, q = 30º, 150º

Infinity Optional Mathematics Book - 10 235

3. Solve : (0° ≤ q ≤ 180°) (b) tanq = cot5q
(a) sin2q + cosq = 0

Solution :

(a) Here, Sin2q + cosq = 0

or, 2sinq.cosq + cosq= 0

or, cosq (2sinq + 1)= 0

Either, cosq = 0

or, cosq = cos90º

\ q = 90º

or, 2sinq + 1 = 0
–1
or, sinq = 2

or, sinq = – sin30º

or, sinq = sin (180º + 30º), sin (360º – 30º)

\ q = 210º, 330º

But 0º ≤ q ≤ 180º

\ q = 90º

(b) Here, tanq= cot5q

or, tanq= tan (90º – 5q)

or, q= 90º – 5q

or, 6q= 90º

\ q= 15º

4. Solve (0° ≤ q ≤ 360°)

(a) (3sinq – 4) (2sinq + 3 ) = 0

(b) (2secq + 3) (tanq + 3 ) = 0

Solution :

(a) Here, (3sinq – 4) (2sin q + 3 ) = 0

Either, 3sinq – 4 = 0

or, sinq = 4 (Rejected) [since value of sinq and cosq lies from – 1 to +1]
3

or, 2sinq + 3 = 0

or, sinq = –3
2

or, sinq = ­­– sin60º

or, sinq = sin(180º + 60º), sin(360º – 60º)

Hence, q = 240º, 300º

236 Infinity Optional Mathematics Book - 10

(b) (2secq + 3) (tanq + 3 ) = 0

Solution :

Here, (2secq + 3) (tanq + 3 ) = 0

Either, 2secq + 3 = 0

or, secq = –3
2
\ q = sec–1 –3
2

or, tanq + 3 = 0

or, tanq= – 3

or tanq= – tan60º

or, tanq= tan (180º – 60º), tan (360º – 60º)

\ q= 120º, 300º

Hence, q = sec–1 –3 , 120º, 300º
2
5. Solve (0° ≤ q ≤ 360°)

(a) 2sin2q + 3cosq = 3 (b) tan2q – 3secq + 3 = 0

(c) secq . tanq = 2 (d) 3 sinq + cosq = 2

(e) cosq – 3 sinq = 2 (f) sin2q + sin4q = cosq + cos3q
Solution :

(a) 2sin2q + 3cosq = 3

Here, 2sin2q + 3cosq = 3
or, 2(1 – cos2q) + 3cosq – 3 = 0
or, 2 – 2cos2q + 3cosq – 3 = 0
or, 2cos2q – 3cosq +1 = 0
or, 2cos2q – 2cosq – cosq + 1 = 0

or, 2cosq (cosq – 1) – 1 (cosq – 1) = 0

or, (cosq – 1) (2cosq – 1) = 0

Either, cosq – 1 = 0

or, cosq = 1

or, cosq = cos0º, cos (360º – 0º)

\ q = 0º, 360º

or, 2cosq – 1 = 0
1
or, cosq = 2

or, cosq = cos60º, cos (360º – 60º)

\ q = 60º, 300º

Hence, q = 0º, 60º, 300º, 360º

Infinity Optional Mathematics Book - 10 237

(b) tan2q – 3secq + 3 = 0

Here, tan2q – 3secq + 3 = 0

or, sec2q – 1 – 3secq + 3 = 0

or, sec2q – 3secq + 2 = 0

or, sec2q – 2secq – secq + 2 = 0

or, secq (secq – 2) –1 (secq –2) = 0

or (secq – 1) (secq – 2) = 0

Either, secq – 1 = 0

or, secq = 1

or, secq = sec0º, sec (360º – 0º)

\ q = 0º, 360º

or, secq – 2 = 0

or, secq = 2

or, secq = sec 60º, sec (360º – 60º)

\ q = 60º, 300º

Hence, q = 0º, 60º, 300º, 360º
(c) secq . tanq = 2

Here, secq.tanq = 2

or, co1sθ . sinθ = 2
cosθ

or, sinq = 2 (1 – sin2q)

or, 2 sin2q + sinq – 2 = 0

or, 2 sin2q + 2sinq – sinq – 2 =0

or 2 sinq (sinq + 2 ) – 1 (sinq + 2 ) = 0

or (sinq + 2 ) ( 2 sinq – 1) = 0

Either, sinq + 2 = 0

or, sinq = – 2 (rejected)

OR, 2 sinq – 1 = 0

or, sinq = 1
2

or, sinq = sin45º, sin(180º – 45º)

\ q = 45º, 135º

Hence, q = 45º, 135º

238 Infinity Optional Mathematics Book - 10

(d) 3 sinq + cosq = 2

Here, 3 sinq + cosq = 2 … (i)
Here, (coefficient of sinθ)2 + (coefficient of cosθ)2
= ( 3)2 + (1)2 = 3 + 1 = 2

Now, dividing equation (i) on both sides by 2, we get

sinq . 3 + cosq . 1 = 2
2 2 2
or, sinq . cos30º + cosq . sin30º = 1
2

or, sin (q + 30º) = sin45º, sin(180 – 45)

or, q + 30º = 45º, 135º

\ q = 15º, 105º

Hence, q = 15º, 105º
(e) cosq – 3 sinq = 2

Here, Cos q – 3 sinq = 2

3 sinq – cosq = – 2 … (i)

Here, (coefficient of sinθ)2 + (coefficient of cosθ)2
= ( 3)2 + (-1)2 = 3 + 1 = 2

Now, dividing both sides of equation (i) by 2, we get

3 sinq – cosq = –2
2 2 2

or, sinq . cos30º – cosq . sin30º = –1

or, sin (q – 30º) = ­– sin90º

or, sin (q – 30º) = sin (360º – 90º)

or, q – 30º = 270º

\ q = 300º

Hence, q = 300º
(f) sin2q + sin4q = cosq + cos3q

Here, sin2q + sin4q = cosq + cos3q

or, 2sin 2q + 4q . cos 4q – 2q = 2cos 3q + q . cos 3q – q
22 22

or, 2sin 3q . cosq = 2cos2q . cosq

or, 2sin3q . cosq – 2cos2q . cosq = 0

or, 2cosq (sin3q – cos2q) = 0

Either, 2cosq = 0

Infinity Optional Mathematics Book - 10 239

or, cosq = 0
or, cosq = cos90º, cos (360º – 90º)
\ q = 90º, 270º
or, sin3q – cos2q = 0
or, sin3q = cos2q
or, sin3q = sin (90°– 2q)
or, 3q = 90° – 2q
or, 5q = 90º
\ q = 18º
Hence, q = 18º, 90º, 270º



Exercise 5.4

1. Find the value of q. (0° ≤ q ≤ 360°) (c) 3 tanq = 1
(a) sinq = 21 (b) 2 cosq = 1
(f) 2sinq + 2 = 0
(d) 2sinq + 3 = 0 (e) 3cotq – 3 = 0 (i) 3sec2q – 4 = 0
(l) 4cosec2q – 3 = 5
(g) 5tanq + 5 = 0 (h) 4sin2q – 3 = 0

(j) 4cos2q – 1 = 0 (k) 3 tan2q – 9 = 0

2. Find the value of q (0° ≤ q ≤ 180º)

(a) (2sinq – 1) (sinq – 1) = 0 (b) ( 3 tanq – 1) (tanq + 3 ) = 0
(d) (10sinq + 5) (2cosq + 1) = 0
(c) (2cosq + 3 ) ( 2 cosq – 1) = 0 (f) 7sin2q + 3cos2q – 4 = 0
(h) 6cos2q + 4sin2q = 5
(e) sinq (4sin2q – 1) = 0 (j) 4cos2q + 4sinq = 5
(l) 4cos2q – 4cosq + 1 = 0
(g) 4sec2q – 7tan2q = 3 (n) cosecq – 2sinq = 1
(p) tan2q + (1 – 3 ) tanq – 3 = 0
(i) 4sin2q + 2cos2q = 3 (r) secq . tanq = 2

(k) 2 3 cos2q = sinq

(m) tan2q – 3secq + 3 = 0

(o) 2sinq + cotq – cosecq = 0

(q) cot2q + 3 + 1 cotq + 1 = 0
3

240 Infinity Optional Mathematics Book - 10

(s) 3cotq – tanq = 2 (t) 4cos2q2 + 4cosq2 – 3 = 0
3. Solve (0° ≤ q ≤ 180°)

(a) sin3q = cos6q (b) sec7q = cosec2q (c) tan15q = cot3q

(d) cot5q = tanq (e) 4cosq = 3secq (f) 2sinq = cosecq

(g) tanq + cotq = 2 (h) sinq + sin2q = 0 (i) sinq – cosq = 0

(j) 2 sinq = tanq (k) 1+ cos2q = cosq (l) sinq – tanq = 0

(m) cosq (2sinq – 1) = 0 (n) sinq . cos45º + cosq . sin45º = 1
2

(o) cosq . cos30º – sinq . sin30º = 21 (p) sinq.cos30º – cosq.sin30º = 1

4. Solve (0 ≤ q ≤ 360°)

(a) sinq + cosq = 1 (b) 3 sinq + cosq = 2

(c) sinq + 3 cosq = 1 (d) 3 sinq – cosq = 3

(e) cosq – sinq = 1 (f) cosq – 1 sinq = 1
2 3

(g) 3 sinq – cosq = 2 (h) secq – 3 tanq = 1

(i) tanq + 3 secq = 3 (j) cosecq + cotq = 3
(k) cotq – cosecq = ­– 3 (l) 3 cos2q + sin2q = 2

5. Solve the following equation : (0º ≤ q ≤ 360º)

(a) sin4q + sin2q = cosq (b) cos3q + cosq = cos2q

(c) sin3q + sinq = 2sinq (d) cos3q + cosq = 2cosq

(e) cos3q + cos2q + cosq = 0 (f) cosq + sinq = cos2q + sin2q

(g) sin2q + sin4q = cosq + cos3q

Infinity Optional Mathematics Book - 10 241

5.6 Height and Distance

As we know that trigonometry is the branch of mathematics that deals with the

measurement of different parts of a triangle by means of a right angled triangle.

It is used not only to measure the sides and angles of a triangle, but it is used to

measure the distance between two points, height of objects, breadth of the river etc

which are impossible to measure practically. Thus when the actual measurement

of the height of an object or distance between two points is not easy or even not

possible, as an application of a trigonometry, a technique is used to find them with

the help of the angle(S) subtended at a point by the object (S) whose distance or

height is to be determined. The instruments like theodolites or sextants are used

to measure the angles and then method of solution of triangle is used to find the

required height or distance. This method is mostly used in surveying, map-making,

aviation, navigation and astronomy etc. A (Object)

Angle of elevation Line of sight
In the adjoining figure, O is the observer and A is
the position of the object, A is obviously at a higher θ B
level than O. OA is the line of sight and OB is the O Horizental
horizontal line through the observation point O. Observer
Then ∠AOB is called the angle of elevation.

Hence when an observer observes on object lying at the higher level, the angle

formed by the line of sight with the horizontal ground or a line parallel to the

ground is called an angle of elevation of that point. Angle of elevation is also called

altitude. Observer C
Angle of dispersion O

θ

In the adjoining figure, O is the observer and A is

the position of the object. Here the object A is at Line of sight
a lower level than the observer O. OA is the line
of sight and OC is the line parallel to the ground

AB. Then ∠COA is called the angle of depression.

In the figure, OC and AB are parallels. So ∠COA B A (Object)

=∠OAB = q

242 Infinity Optional Mathematics Book - 10

Thus when an observer observes an object lying in lower level, the angle formed
by the line of sight with horizontal line parallel to the ground is called angle of
depression or angle of declination.

WORKED OUT EXAMPLES

1. Two men are on the opposite side of a tower of 80m high. They
observed the angle of elevation of the top of the tower and found to
be 30º and 60º. Find the distance between them.

Solution:

Here, AD = 80m is the height of tower. ∠ABD = 30º and ∠ACD = 60º are the

angles of elevation made by two men to the top of the tower and BC = distance

between the men = ? A

In right angled Δ ABD.

Tan 30º = AD 80 m
BD
or, 13 30° 60°
= 80 B D C
BD

\ BD = 138.56m.

Again, in right angled ΔADC

Tan 60º = AD
DC

or, 3 = 80
DC

\ DC = 46.18m.

Now, BC = BD + DC

= 138.56m + 46.18m

= 184.75m

Hence, distance between the two men (BC) = 184.75m.

2. The angles of elevation of the top of a tower at two places due east
of its foot are 45º and 30º. If the two places are 180m apart, find the
height of the tower.

Solution:

Here, AB = height of tower = ?

DC = 180m is the distance between two points of observation.

Infinity Optional Mathematics Book - 10 243

∠ADB = 45º and ∠ACB = 30º are the angles of elevation of the top of the tower.

Now, In right angled Δ ABD, N
WE
Tan 45º = AB A
BD ? S

or, 1 = AB
BD

\ BD = AB. 45° 30°
B DC
Again, in right angled ΔABC
←180 m→
AB
Tan 30º = BC

or, 1 = AB
3 BD + DC

or, 1 = AB [∵ BD = AB]
3 AB + 180

or, AB + 180 = 3 AB

\ AB = 245.9 m.

Hence, height of tower (AB) = 245.9 m.

3. A flagstaff is fixed on the top of a building. Find the length of the

flagstaff if the angles of elevation of the tops of the flagstaff and the

building are observed from a point at a distance of 25m from the

building are 60° and 45º respectively. A

Solution:
Here, BC = 25m is the distance between the observer and the ?

building D

AD = height of flagstaff = ?, BD = height of building

∠DCB = 45º and ∠ACB = 60º are the angles of elevation.

Now, in right angled ΔDCB. 60° 45°
B 25 m C
Tan 45º = BD
BC

or, 1 = BD
25

\ BD = 25m.

Again, in right angled ΔABC

Tan 60º = AB
BC

or, 3 = AD + BD
25

or, 25 3 = AD + 25

\ AD = 18.25 m.

Hence, length of the flagstaff (AD) = 18.25m.

244 Infinity Optional Mathematics Book - 10

4. From the top of a cliff, the measure of the angle of depression of the
top and the bottom of a building are found to be 30º and 45º. If the
height of the cliff is 100m, find the height of the building.

Solution: Here, AB = 100 m is the height of cliff. ∠FAD = 30° and ∠FAC = 45° are

the angles of depression of the top and bottom of the building from the top of

cliff. A 30° 45° F
DC = height of building = ?

In right angled DABC, 100 m E 30° D

tan 45° = AB ?
BC 45°
BC
or, 1= 100
BC

\ BC = 100m

Here, BC = ED = 100 m

Again, In right angled DAED,

tan 30° = AE
CD
1
or, 3 = AE
100

\ AE = 57.73 m

Now, DC = EB = AB – AE = (100 – 57.73) m = 42.27 m

Hence, height of building (DC) = 42.27 m

5. The angle of depression and elevation of the top of a pole 25m high
from the top and bottom of the tower are 60° and 30° respectively.
Find the height of the tower.

Solution: Here, AD = Height of the tower = ?, CE = 25m is the height of pole.

∠FAC = 60º is the angle of depression and ∠CDE = 30º is the angles of
elevation from the top and bottom of the tower to the top of pole.

Now, In right angled ΔCDE. A 60° F

Tan 30º = CE
DE

or, 1 = 25 ? B 60° C
3 DE

\ DE = 43.3m 25 m

Here, DE = BC = 43.3m 30°
D
E

Infinity Optional Mathematics Book - 10 245

Again, in right angled ΔABC

Tan 60º = AB
BC

or, 3 = AB
43.3

or, AB = 75m

Here, CE = BD = 25m

AD = AB + BD = (75 + 25)m = 100m.

Hence height of the tower (AD) = 100m.

6. There are two posts of height 90m and 30m respectively. From the
foot of the second, the elevation of the top of the first is found to be
60º. Find the angle of elevation of the top of the second from the foot
of the first.

Solution:

Here, AB = 90m and CD = 30m are the heights of two posts.

∠BDA = 60º is the angle of elevation from the foot of second to the top of the

first post.

& let ∠CBD = q is the angles of elevation from the foot of the first to the top of

the second post. A
Now, In right angled Δ ABD.

Tan 30º = BADB

or, 3 = 90 C
BD 90 m
30 m
or, BD = 90 = 30 3m
3

Again, In right angled ΔBCD q=? 60°
B D
tanq = 30
30 3

or, tanq = 1
3
or, tanq = tan30º

\ q = 30º

Hence, ∠CBD = 30º

Which is the required angle of elevation.

7. Two posts are 180m apart and height of one is double the height of other. From the
midpoint of the line joining their feet, an observer finds the angle of elevation of their
tops to be complementary. Find the height of the longer post.

Solution:

Here, DE = x m. is the height of shorter post

246 Infinity Optional Mathematics Book - 10

AB = 2xm is the height of longer post.

BE = 180m is the distance between two posts.

∠ACB = q and ∠DCE = (90° – q) are the angles of elevation.

Here, BC = CE = 90m A

Now, In right angled ΔCDE D
DE
Tan (90° – q) = CE ?
or, cotθ x θ 90° – θ
= 90 … (i) 2x m B CE

Again, In right angled ΔABC, 180 m
AB
tanθ = BC

or, tanθ = 2x … (ii)
90
Now, multiplying (i) and (ii) we get
x 2x
cotθ . tanθ = 90 × 90

or, 2x2 = 8100

or, x2 = 4050

∴ x = 63.63m

Now, AB = 2x = 2 × 63.63 = 127.27m

Hence height of longer post (AB) = 127.27m.

8. AB is a vertical pole with its foot B on a level ground. C is a point
on AB such that AC:CB = 3:2. If the parts AC and CB subtend equal
angles at a point on the ground which is at a distance of 20m from the
foot of the pole, find the height of the pole.

Solution: A

Here, AC:CB = 3:2

Let AC = 3xm and CB = 2xm 3x

∠ADC = ∠CDB = q C

Now, In right angled ΔCBD 2x θ D
B θ
CB 2x x
Tanq = BD = 20 = 10 ………. (1) 20 m

Also, In right angled ΔABD,

tan2q = AB
BD
2tanq 5x
or, 1 – tan2q = 20

2 × x x
– 10 4
or, x 2 =
1 10

Infinity Optional Mathematics Book - 10 247

or, 10020– x2 = 1
4

or, 100 – x2 = 80

or, x2 = 20
or, x = 2 5m

Hence, height of the pole (AB) = 5x = 5 × 2 5m = 22.36m

9. A ladder 10m long reaches a point 10m below the top of a vertical flagstaff. From
the foot of the ladder, the angle of elevation of the top of the flagstaff is 60°. Find the
height of flagstaff.

Solution: A
Here, BD = 10m is the length of ladder

In the figure, ∠CAD = 30º 10 m

Also, ∠BAD = ∠ADB = 30º [∠AB = BD]

B 10 m

Then, ∠BDC = ∠ADC – ∠ADB = 60º – 30º = 30º 60° D
C
Now, In right angled ΔBCD

Sin30º = BC
BD

or, 1 = BC
2 10

\ BC = 5m

Hence, height of flagstaff (AC) = AB + BC

= (10 + 5)m

= 15m.

10. The angle of elevation of a bird from a point 400 ft above a pond is 30°. From the
same point, the angle of depression of its image in the pond is found to be 60°. Find
the height of the bird from the water surface.

Solution:

Here, AE = height of the bird from the water surface.

∠ACB = 30º is the angle of elevation.

∠BCA' = 60º is the angle of depression of the image A' of the bird.

Now, In right angled ΔABC

tan30º = AB
BC

or, 1 = AB
3 BC

248 Infinity Optional Mathematics Book - 10

or, BC = 3 (AE – 400) … (1) A
B?
Again, In right angled ΔBCA' C 30° E
400 Dft 60°
BA' A'
Tan 60º = BC

or, 3 = BE + EA'
3 (AE – 400)

or, 3(AE – 400) = BE + AE [∴ AE = A'E]

or, 2AE = 1200 + 400

\ AE = 800ft.

Hence, height of the bird from the water surface = 800ft.

Exercise 5.6

1. (a) A man of height 1.5m observes the angle of elevation of the top of the
pole situated infront of him and finds to be 60º. If the height of the pole is
121.5m, find the distance between the pole and the man.

(b) From the top of the building the angle of elevation of the top of the tower is
found to be 30º. The distance between the building and the tower is 120m.
Find the height of the building, if the height of the tower is 100m.

(c) From the top of the building the angle of depression of the top of the pole
having height 10m is 60º. If the distance between the building and the pole
is 50m, find the height of the building.

(d) The upper part of a tree broken by the wind makes an angle of 60º with the
horizontal ground at a distance of 15m from the foot of the tree. Find the
height of the tree before it was broken.

(e) The upper part of a tree broken by the wind is 20m long. It touches the
ground at the distance of 10m from the bottom of the tree. Find the angle
made by the broken part of the tree with the ground and the height of the
tree before it was broken.

2. (a) The angle of elevation of the top of the tower of height 100m from the
opposite side of the tower are found to be 45º and 60º respectively. Find the
distance between the two points of observation on the ground.

(b) Two students are on the opposite side of a house of height 100m. They
observed the angle of elevation of the top of the tower and found to be 30º
and 60º. Find the distance between them.

(c) A man on the top of a tower of height 100m, observes the angles of
depression of two houses on the opposite sides of the tower to be 30º and
45º. Find the distance between the two houses.

Infinity Optional Mathematics Book - 10 249

(d) The angles of elevation of the balloon flying 200m high on the sky from the
two places on the opposite sides are 45º and 60º. Find the distance between
the two places.

(e) A cloud is observed above a river from opposite bank and the angles of
elevation are found to be 45º and 60º. The cloud is vertically above the line
joining the points of observation and the river is 80m wide. Find the height
of the cloud.

3. (a) A man observes the angle of elevation of the top of tower from a place to be
30º. On walking 50m towards the tower, the angle of elevation is found to
be 45º. Find the height of the tower.

(b) The angles of elevation of the top of a school building from the two places
on the same side are 60º and 45º. If the distance between two places is
100m, find the height of the school building.

(c) The angle of elevation of a tower observed from a point is 45º. On walking
200m away from that point the angle changes to 30º. Find the height of the
tower.

(d) From the top of a tower 100m high, the angles of depression of two places
due east are 60º and 45º. What is the distance between two places ?

(e) From an aeroplane, vertically above a straight road, the angles of
depression of two places on the road are observed to be 30º and 45º. If the
distance between the two places is 600m, find the height of the aeroplane.

(f) From the top of a tower, a man observes that the angles of depression of
two consecutive km stones in a straight road in front of him are 60º and 45º
respectively. Find the height of the tower.

(g) From the top of a cliff 200m high, the angles of depression of two boats due
west of the cliff are 45º and 30º. Find the distance between the two boats.

(h) From the top of a tower 192ft high, the angle of depression of two vehicles

on a road at the same level as the base of the tower and on the same side
3 13. Calculate the distance
of it are xº and y° where tan xº = 4 and tan yº =
between the vehicles.

(i) The shadow of a tower on the ground is found to be 45m longer when the
sun's altitude is 45º than when it is 60º. Find the height of the tower.

(j) The shadow of a tower standing on a level plane is found to be 60m longer
when the sun's altitude is 30º then when it is 45º. Prove that the height of
the tower is 30(1 + 3 )m.

(k) The shadow of a tower on the level ground increases in length by y metres
when the altitude changes from 45º to 30º. Calculate the value of y, given
that the height of the tower is 25metres.

(l) From a point on the bank of river, it is observed that the angle subtended
by a house on the opposite bank is 45º, By retiring 50m from the bank, the

250 Infinity Optional Mathematics Book - 10