Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

(VIII) cosD - cosC = 2sin C + D. sin C-D .
2 2
C + D. D-C
or, cosC - cosD = 2 sin 2 sin 2

Above formulae (i) to (viii) are the transformation formulae of trigonometric ratios.

Worked out Examples

Example 1. Express the following product forms into sum or difference forms:
Solution:
(a) sin24° cos12° (b) cos42° sin22°
Example 2.
Solution: (c) cos50°.cos35° (d) sin40° sin20°

(a) sin24°.cos12° = 1 (2sin24°cos12°)
2
1
= 2 [sin(24° + 12°) + sin(24° - 12°)]

= 1 [sin36° + sin12°]
2
1
(b) cos42°. sin22° = 2 [2cos42°.sin22°]

= 1 [sin(42° + 22°) - sin(42° - 22°)]
2
1
= 2 [sin64° - sin20°]

(c) cos50° cos35° = 1 [2cos50° cos35°]
2
1
= 2 [cos(50° + 35°) + cos(55° - 35°)]

= 1 [cos85° + cos20°]
2
1
(d) sin40°. sin20° = 2 [2sin40°.sin20°]

1 [cos(40° - 20°)- cos(40° + 20°)]
(= 2
1 1 1
= 2 [cos20° - cos60°] = 2 cos20°- 2 (

Express the following sum or difference into product form:

(a) cos6T + cos4T (b) sin50° - sin40°

(c) sin50° - sin40° (d) cos35° - cos25°

(a) cos6T + cos4T = 2cos 6T + 4T . cos 6T - 4T
2 2
= 2cos5T. cosT

(b) sin50° + sin40° = 2sin 50° + 40° . cos 50° - 40°
2 2
= 2sin45°.cos5°

(c) sin50° - sin40° = 2cos 50° + 40° . sin 50° - 40°
2 2
= 2cos45°. sin5°

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(d) cos35° - cos25° = 2sin 35° + 25° . sin 25° - 35°
2 2
= 2sin30°.sin(-5°)

= –2sin30°.sin5° [ sin(-T) = - sinT@

= –2 . 1 . sin15° = – sin15°
2

Example 3. Prove that
Solutions:
(a) sin75° + sin15° = 3 (b) sin18° + cos18° = 2 cos27°
2
1
(c) sin(45° - A) sin (45° + A) = 2 cos2A

(a) LHS = sin75° + sin15°

= 2sin 75° + 15° . cos 75° - 15°
2 2
= 2sin45°. cos30°

=2. 1 . 3 = 3 = RHS Proved.
2 2 2

(b) LHS = sin 18° + cos18° = sin18° + cos(90° - 72°)

= sin18° + sin72°

( ) ( )= 2sin
18° + 72° . cos 72° - 18°
2 2
1
= 2sin45°. cos27° = 2. 2 . cos27°

= 2 cos27° = RHS. Proved.

(c) LHS = sin (45° - A). sin (45° + A)

= 1 [2sin(45° - A). sin(45° + A)]
2
1
= 2 [cos(45° - A - 45° - A) - cos(45° - A + 45° + A)]

= 1 [cos (-2A)- cos90°] = 1 cos2A = RHS. Proved.
2 2

Example 4. Proved that
Solution:
(a) sin4A + sin2A = tan3A (b) cos8° + sin8° = tan53°
cos4A + cos2A cos8° – sin8°

(a) LHS = sin4A + sin2A = tan3A
cos4A + cos2A
( ) ( )2sin
4A + 2A . cos 4A - 2A = sin3A . cosA
= 2 . cos 2 cos3A. cosA

( ) ( )2cos4A + 2A 4A - 2A
2 2
= tan3A = RHS Proved

(b) LHS = cos8° + sin8°
cos8° – sin8°
cos8° + sin(90° - 82°) cos8° + cos82°
= cos8° - sin(90° - 82°) = cos8° - cos82°

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(( )) (( ))=2cos8 + 82° . cos 8°-82° = sin45°. cos37°
2sin 2 . sin 2 sin45°.sin37°

cos37° 8° + 82° 82° - 8
sin37° 2 2

= = cot 37°

= cot (90° - 53°) = tan53° = RHS. Proved.

Example 5. Proved that sin7T - sin5T - sin3T + sinT = tan2T
Solution: cos7T + cos3T - cos5T - cosT

Example 6. LHS = sin7T - sin5T - sin3T + sinT
Solution: cos7T + cos3T - cos5T - cosT

Example 7. = (sin7T + sinT) - (sin5T + sin3T)
Solution: (cos7T - cosT) + (cos3T - cos5T)

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

= sin4T . cos3T - sin4T . cosT = sin4T (cos3T - cosT)
sin4T . sin(–3T) + sin4T .sinT sin4T (sinT - sin3T)

2sin 3T + T . sin T - 3T
2 2
sin2T
= 2cos T + 3T . sin T - 3T = cos2T = tan2T = RHS. Proved.
2 2

Prove that sin8T . cosT - sin6T . cos3T = tan2T
cos2T . cosT - sin3T . sin4T
sin8T . cosT - sin6T . cos3T
LHS = cos2T . cosT - sin3T . sin4T

= 2sin8T . cosT - 2sin6T . cos3T
2cos2T . cosT - 2sin3T . sin4T

= sin(8T + T) + sin(8T - T) - sin(6T + 3T) - sin(6T - 3T)
cos(2T + T) + cos(2T - T) - cos(3T - 4T) + cos(3T + 4T)

= sin9T + sin7T - sin9T - sin3T = sin7T - sin3T
cos3T + cosT - cosT + cos7T cos3T + cos7T

= 2cos 7T +3T . sin T - 3T = sin2T
2 2 cos2T

2cos T + 7T . cos 3T - 7T
2 2
= tan2T = RHS. Proved.

Prove that sin20°.sin40°.sin60°.sin80°.Sun120° = 33
32
LHS = sin20°.sin40°.sin60°.sin80°.sin120°

= sin20° sin40° 3 sin80° 3
2 2
3
= 4×2 sin20° [2sin40°. sin80°]

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= 3 . sin20° [cos(40° - 80°) - cos(40° + 80°)]
8
3
= 8 sin20° [cos40° - cos120°]

= 3 sin20° .cos40° - 3 sin20° (– 1 )
8 8 2
3 3
= 8 sin20° .cos40° + 16 sin20°

= 3 [sin(20° + 40°) + sin(20° - 40°)] + 3 sin20°
16 16
3 3
= 16 [sin60° + sin(-20°)] + 16 sin20°

= 3 . 3 - 3 sin20° + 3 sin20° = 33 = RHS. Proved.
16 2 16 16 32
1
Example 8. Prove that cos20°.cos40°. cos60°.cos80° = 16 .
Solution:
LHS = cos20°. cos40°.cos60°.cos80°
Example 9.
Solution: = cos20°.cos40° 1 . cos80°
2
228 1
= 4 cos20° (2cos40°.cos80°)

= 1 cos20° [cos(40° + 80°) + cos(40° - 80°)]
4
1
(= 4
cos20° [cos120° + cos40°]

= 1 cos20° - (+ 1 cos20°.cos40°
4 ( 4
1 1
= – 8 cos20° + 8 (2cos20° . cos40°)

= – 1 cos20° + 1 [cos(20° + 40°) + cos(20° - 40)]
8 8
1 1 1 1 1 1
= – 8 cos20° + 8 cos60° + 8 cos20° = 8 . 2 = 16 = RHS. Proved.

Alternative Method

LHS = cos20°. cos40°.cos60°.cos80°

= 1 (2 sin20°cos20° ) cos40° 1 cos80°
2sin20° 2
1
= 4sin20° sin40°cos40° cos80°

= 1 (2sin40°cos40°).cos80° = 1 (2sin80°cos80°)
8sin20° 16sin20°
1 1 1
= 16sin20° sin160° = 16sin20° . sin20° = 16 = RHS Proved.

Prove that sinT.sin(60° - T) sin(60° + T) = 1 sin3T
4
LHS = sinT.sin(60° - T) sin(60° + T)

= 1 sinT [2sin(60° - T). sin (60° + T)]
2
1
= 2 sinT [cos(60° - T - 60 - T) - cos(60° - T + 60° + T)]

(= 1 [cos
2 sinT [-2T] - cos120°]

= 1 sinT.cos2T - 1 . sinT -
2 2

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vedanta Excel In Opt. Mathematics - Book 10

= 1 [sin (T + 2T) + sin (T - 2T)] + 1 sinT
4 4
1 1 1 1
= 4 sin3T - 4 sinT + 4 sinT = 4 sin3T = RHS. Proved.

Example 10. Prove that cos Sc . cos 2Sc . cos 4Sc . cos 7Sc = 1
Solution: 15 15 15 15 16
Sc 2Sc 4Sc 7Sc
Example 11. (LHS=cos 15 , cos 15 , cos 15 . cos 15
Solution:
Sc Sc cos 2Sc . cos 4Sc . cos 7Sc
15
2sin .cos ((
((
= 2sin Sc
( 15

= sin 2Sc . cos 2Sc ( . cos 4Sc . cos 7Sc
(= 15 15 ( 15 15

2sin 2Sc .cos (2Sc . cos 4Sc . cos 7Sc
15 (
(
4sin Sc
(= 15 (
1
2sin 4Sc .cos 4Sc . cos 7Sc
8sin Sc 15
(= 15
1
sin 8Sc . cos cos 7Sc
8sin Sc
(= 15
1
sin Sc - 7Sc .cos 7Sc
8sin Sc
(= 15
1
2sin 7Sc cos 7Sc
16sin Sc
15
1
16sin sin 14Sc
(= Sc
15
1
= 16sin Sc . sin Sc - Sc
15

= 1 . sin Sc = 1 = RHS. Proved.

16sin Sc
15

Prove that 2cos Sc . cos91S3c + cos31S3c + cos51S3c = 0
13
Sc cos91S3c 3Sc cos51S3c
2cos 13 13 +
( (LHS= . + cos

= cos Sc + 9Sc + cos Sc – 9Sc + 3Sc + cos 5Sc
13 13 13 13 13 13
10Sc 8Sc 3Sc 5Sc
13 13 13 13
( (= cos + cos + cos + cos

= cos Sc – 3Sc + cos Sc – 5Sc + cos 3Sc + cos 3Sc
13 13 13 13
3Sc 5Sc 3Sc 5Sc
= - cos 13 - cos 13 + cos 13 + cos 13 = 0 = RHS Proved.

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Example 12. (HIfesrei1n, Dsi1n+Dc+o1sDco=1sDsi=1nEs+i1nEco1+sEprove that cot D+E (, = tanD.tanE.
Solution: 1 2
cosE
1 1 1 1
or, sinD - sinE = cosE - cosD

or, sinE - sinD = cosD - cosE
sinD.sinE cosD.cosE
+ E-D E-D
2cos E 2 D . sin 2 2sin D E . sin 2
2
or, =
((or, sinD . sinE cosD . cosE

cos D+E ( sinD . sinE
2 ( cosD.cosE
=
D+E (
sin 2 (
(? (
cot D+E = tanD.tanE Proved.
2

Example 13. Prove that sin2D - sin2E = tan(D + E)
sinD.cosD - sinE.cosE
sin2D - sin2E 2sin2D - 2sin2E
Solution: LHS = sinD.cosD - sinE.cosE = 2sinD.cosD - 2sinE.cosE

1-cos2D - 1 + cos2E cos2E - cos2D
sin2D - sin2E sin2D - sin2E
( (= =

2sin 2E + 2D . sin 2D - 2E (
( (= 2 2 ( sin(D + E)
2cos 2D + 2E . sin 2D - 2E = cos(D + E)

22
= tan (D + E) = RHS. Proved.

Example 14. If sin2x + sin2y = 1 and cos2x + cos2y = 1 , then show that
2 3 2
tan (x + y) = 3 .

Solution: ( (Here, sin2x + sin2y = 1
3
1
2x + 2y 2x - 2y 3
2
or, 2sin (( . cos 2 =
or, (( ............. (i)
2sin (x + y).cos (x - y) = 1
3
1
( (Again, cos2x + cos2y = 2 1
2
2cos 2x + 2y . cos 2x - 2y =
or, 2 + y). cos(x y) =221
- ............. (ii)
2cos(x

Dividing (i) by (ii), we get,

2sin(x + y). cos(x - y) 1
2cos(x + y).cos(x - y) 3
= 1

tan (x + y) = 2 Proved. 2
3
?

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vedanta Excel In Opt. Mathematics - Book 10

Exercise 9.3

Very Short Questions

1. Express each of the following as sum or difference form:

(a) 2sinA. sinB (b) 2cosA.cosB (c) 2cosA.sinB (d) 2sinA.cosB

(e) 2sin5T. cos3T (f) 2sin3x. sinx (g) 2sin20°.sin10° (h) 2cos40°.cos20°

2. Express each of the following as product form:

(a) sin6T + sin2T (b) sinx - siny (c) cosx - cosy

(d) sinx + siny (e) sin50° + sin40° (f) cos70° - cos40°

(g) cos40° - sin20° (h) sin5x - sin3x (i) cos7x - cos11x

(j) cos5x + cos2x

3. Prove the following:

(a) cos75° + cos15° = 3 (b) sin75° - sin15° = 1
2 2
(d) sin50° + sin70° = 3 cos10°
(c) sin18° + cos18° = 2 cos27°

(e) sin50° + sin10° = sin70° (f) cos52° + cos68° + cos172° = 0

Short Questions

4. Prove the following :

(a) cos5A + cos3A = cotA (b) cos40° - cos60° = tan50°
sin5A - sin3A sin60° - sin40°
cos80° + cos20° cos8° + sin8°
(c) sin80° - sin20° = 3 (d) cos8° - sin8° = tan53°

(e) cos10° - sin10° = cot55° (f) cos(40° + A) + cos(40° - A) = cotA
cos10°+ sin10° sin(40° + A) - sin(40° - A)

Long Questions

5. Prove the following :

(a) sinA.sin2A + sin3A.sin6A = tan5A
sinA.cos2A + sin3A.cos6A
cos2A.cos3A - cos2A.cos7A sin7A + sin3A
(b) sin4A.sin3A - sin2A.sin5A = sinA

(c) sinA + sin3A + sin5A + sin7A = tan4A
cosA + cos3A + cos5A + cos7A
cos7A + cos3A - cos5A -cosA
(d) sin7A - sin3A - sin5A + sinA = cot2A

(e) sin5A - sin7A - sin4A + sin8A = cot6A
cos4A - cos5A - cos8A + cos7A
sin(p+2)T - sinpT
(f) cospT - cos(p + 2) T = cot (p + 1) T

(g) (sin4A + sin2A) (cos4A - cos8A) =1
(sin7A + sin5A) (cosA - cos5A)

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6. Prove the following :

(a) ssccsiiioonnnss2124200000°°°°°.....cssscoiiionnnss4434100000°°°°0....css°siio.inncsno8665s00001°°°°.6.=.sc0ison°isn8=8837000°°18°===136111616

(b)
(c)
(d)
(e)

(f) tan20° tan40° tan80° = 3

7. Prove the following :

(a) sin(45° + T). sin(45° - T) = 1 cos2T
2
1
(b) cos(45° + T). cos(45° - T) = 2 cos2T
(c) cosT. cos(60°
– T) . cos(60 +T) = 1 cos3T
4
1
(d) sinx. sin(60° – x). sin(60° + x) = 4 sin3x

8. Prove that cos(36° - T).cos(36° + T) + cos(54° + T). cos(54° - T) = cos2T

((a)
9. Prove the following :

If 1 – 1 =- 1 – co1sB, prove that cot A+B ( = – tanA.tanB
sinA cosA sinB 2 (
((b)
If 4 + 2 = 2 + 4 , prove that tan D+T = 1
secT secD cosecD secD 2 2

((a)
10. Prove the following : A-B (
2 (
(cosA + cosB)2 + (sinA + sinB)2 = 4cos2
((b)
( ((c) = 4sin2 A -B
2
(d) 3
(cosB - cosA)2 + (sinA - sinB)2 2
sin2
sin2A Sc + A (( Sc A 1 sinA
8 2 - sin2 8 - 2 = 2

+ sin2 (A + 120°) + sin2 (A - 120°) =
(11. (a)
1 1 D+E 1
4 2 2 2
If sinD+ sinE= and cosD+cosE = , then(( prove that : tan =
(
(
((b) 1 1 D+E 3
If cosD + cosE = 3 and sinD + sinE = 4 , prove that : tan 2 = 4
( ((c) If sinx = k siny, prove that : tan
x-y = k-1 tan x+y
2 k +1 2
(d) If sin (A + B) = k sin (A - B), then prove that : (k - 1) tanA = (k + 1) tanB.

12. Prove that sin2x - sin2y = tan(x + y)
sinx.cosx - siny.cosy
1
(13. Prove that cos 2Sc + cos 4Sc + cos 6Sc = - 2
7 7 7
D+E (
14. Prove that x = y cot 2 if xcosD + ysinD = xcosE + ysinE

15. (a) Prove that 1 – cos10° + cos40° – cos50° = tan5°.cot20°
1 + cos10° – cos40° – cos50°

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(b) Prove that 1 – cosT + cosD – cos(D T) = tan T .cot D
1 + cosT – cosD – cos(D + T) 2 2

1. (a) cos(A - B) - cos(A + B) (b) cos(A + B) + cos(A - B)

(c) sin(A + B) - sin(A - B) (d) sin(A + B) + sin(A - B)

(e) sin8T + sin2T (f) cos2x - cos4x (g) cos10° - 3 (h) 1 + cos20°
2 2
x-y x+y
2. (a) 2sin4T . cos2T (b) 2sin 2 . cos 2

(c) 2sin x+y . sin y-x (d) 2sin x+y . cos x-y
2 2 2 2

(e) 2 cos5° (f) -2sin15° . sin55°

(g) 2sin55° . sin15° (h) 2cos4x . sinx

(i) 2sin9x . sin2x (j) 2cos 7x . cos 3x
2 2

9.4 Conditional Trigonometric identities

Identities which are true under given certain conditions are known as conditional identities.
In this section, we deal with some trigonometric identities like A + B + C = Sc,
then A + B = Sc - C, B + C = Sc - A, C + A = Sc - B
We can write,
(i) sin (A + B) = sin (Sc - C) = sinC

sin(B + C) = sin (Sc - A) = sinA
sin(C + A) = sin (Sc - B) = sinB
(ii) cos (A + B) = cos(Sc - C) = - cosC
cos (B +C) = cos (Sc - A) = - cosA
cos(C + A) = cos (Sc - B) = - cosB
(iii) tan(A + B) = tan (Sc - C) = - tanC
tan (B + C) = tan(Sc - A) = - tanA
tan(C + A) = tan (Sc - B) = - tanB

(b) If A + B + C = Sc,

then A + B + C = S2c.
2 2 2
A B C B C A C A B
So, we write 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2
2 2 2

We write,
((i) (= sin
sin A + B (( (= sin Sc - C = cos C
(sin 2 2 (( 2 2 2
B C A A
2 + 2 Sc - 2 = cos 2
2

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(sin C A (= sin Sc B B
((ii) 2 + 2 (= cos (2-2= cos 2

cos A + B (( Sc - C = sin C
(cos 2 2 (((= cos 2 2 2
B C (( A A
(cos 2 + 2 (= cos(( Sc - 2 = sin 2
(( 2
C A (( B B
((iii) tan2 + 2 (= tan(( Sc - 2 = sin 2 .
2
A B C C
(tan 2 + 2 (= tan Sc - 2 = cot 2 ,
2
B C A A
(tan 2 + 2 (= tan Sc - 2 = cot 2
2
C A B B
2 + 2 Sc - 2 = cot 2
2

Worked Out Examples

Example 1. If A + B + C = Sc, prove that
Solution: (a) tanA + tanB + tanC = tanA . tanB . tanC

(b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1
2 2 2
(a) tanA + tanB + tanC = tanA. tanB.tanC

Example 2. Here, A + B + C = Sc.
or, A + B = Sc - C
? tan (A + B) = tan (Sc - C)

or, tanA + tanB = – tanC
1 – tanA.tanB
or, tanA + tanB = – tanC + tanA.tanB.tanC

? tanA + tanB + tanC = tanA. tanB.tanC proved.

(b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1
2 2 2
Here, A + B + C = Sc

or, A + B = Sc – C
2 2 2 2
tan(A2 B2 ) tan(S2c C
? + = – 2 )

tan A + tan B C
2 2 2
(or, A B = cot
1 - tan 2 . tan 2
or,
tan A + tan B = 1 - tanA2 . tan B 1
2 2 2
tan C
2
tanA2 C B C tanA2 B
or, . tan 2 + tan 2 . tan 2 = 1 - . tan 2 .

? tanA2 . tan B + tan B . tan C + tanC2 tanA2 = 1 Proved.
2 2 2

If A + B + C = 180°, then prove the following:
(a) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
(b) cos2A + cos2B + cos2C = - 4cosA.cosB.cosC - 1

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Solution: (c) cos2A + cos2B - cos2C = 1 - 4sinA.sinB. cosC
(a) Here, A + B + C = 180°

sin (A + B) = sin (180° - C) = sinC

and cos(A + B) = cos (180° - C) = - cosC

( (LHS = sin2A + sin2B + sin2C 2A - 2B
2A +2B 2
= 2sin 2 ((. cos + 2sinC.cosC.
((
= 2sin(A + B). cos(A -B) + 2sinC.cosC ((

= 2sinC . cos (A -B) + 2sinC.cosC

= 2sinC[cos(A – B) + cosC]

= 2sinC[cos(A-B)- cos (A + B)]

= 2sinC. 2sinA.sinB

= 4sinA - sinB. sinC = RHS. Proved.

(b) Here, A + B + C = 180°

? cos(A + B) = cos(180° - C) = - cosC

( (LHS = cos2A + cos2B + cos2C 2A - 2B
= 2cos 2A + 2B . cos 2 + 2cos2C - 1
2
= 2cos(A + B). cos(A -B) + 2cos2C - 1

= 2(-cosC). cos (A -B) + 2cos2C - 1

= -2cosC [cos(A-B) - cosC] - 1

=-2cosC. [cos(A - B) + cos(A + B)]- 1

= - 2cosC.[2cosA.cosB]- 1

= -4cosA . cosB . cosC - 1 = RHS. Proved.

(c) Here, A + B + C = 180°

cos(A + B) = cos(180° - C) = -cosC

( (LHS = cos2A + cos2B - cos2C
= 2cos 2A+2B . cos 2A - 2B - 2cos2C + 1
2 2
= 2cos(A + B).cos(A - B) - 2cos2C + 1

= 2cos(A + B).cos(A -B) - 2cos2C + 1

= - 2cosC [cos (A - B) + cosC] + 1

= -2cosC[cos(A - B) - cos(A + B)] + 1

= - 2cosC. 2sinA.sinB + 1

= 1 - 4 sinA. sinB. cosC = RHS. Proved.

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Example 3. If A + B + C = 180°, then prove the following:
Solution:
(a) sinA + sinB + sinC = 4cos A cs.ocinossAA22 B2.. .sciconossB2B2 C2.. ssiinnC2C2
236 (b) cosA + cosB + cosC = 1 + 2
(c) cosA + cosB - cosc = - 1 + 4

4

Here, A + B + C = 180°
( ( ( (Now,
A + B (( C C
2 2 (( 2 2
sin = sin 90° - = cos

and cos A + B = cos 90° - C = sinC2
2 2 2

(a) LHS = sinA + sinB + sinC
( (=
(= A+2- B2sin+C2 s,incoCs
22scions C2A. +2coBs . cos( C
A-B ( 2
(
2((
[ ( ]= 2cosC2
cos A-B + sinC2
2
( ( ]= 2cosC2 [cos
A-B + cos A+B (
2 2
= 2cosC2 2cosA2 B
. . cos 2

= 4 cosA2 . cos B . cos C2 =RHS, Proved.
2

(b) LHS = cosA + cosB + cosC
( (=
(= 2cos A+ B (. cos A-B +1 - 2sin2 C
2 (A-B 2 2sin2 2
C ( C
2sin 2 . cos (( 2 +1- 2
[ ( ]=
2sin C . cos A-B - sin C +1
2 2 2
[ ( ( ]= 2 sinC2
cos A-B - cos A+ B ( +1
2 2
C 2sinA2 B
= 2sin 2 . . sin 2 + 1

= 1 + 4sinA2 . sin B . sin C = RHS. Proved.
2 2

(c) LHS = cosA + cosB - cosC
( (= 2cos
A+ B (. cos A- B - cosC
2 ( 2
(= C (A- B C
2sin 2 .cos (( 2 - 1 + 2sin2 2
[ ( ]= 2 sinC2
cos A-B + sinC2 - 1
2
[ ( ( ]= 2 sinC2
cos A-B + cos A+B ( -1
2 2

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vedanta Excel In Opt. Mathematics - Book 10

= 2sin C . 2cosA2 . cos B - 1
2 2
= - 1 + 4 sinC2 . cosA2 . cosB2 = RHS. Proved.

Example 4. If A + B + C. = 180°, then prove the following:
Solution:
(a) sin2A + sin2B + sin2C = 2 + 2cosA.cosB.cosC

(b) cos2A + cos2B + cos2C = 1 - 2cosA.cosB.cosC.

(c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC

We have, A + B + C = 180°

or, A + B = 180° - C

? sin (A + B) = sin(180° - C) = sinC

and cos(A +B) = cos(180° - C) = -cosC

Now,

(a) LHS = sin2A + sin2B + sin2C

= 1 [2sin2A + 2sin2B] + 1 - cos2C
2
1
= 2 [1 - cos2A + 1 - cos2B] + 1 - cos2C

1 - 1 (cos2A + cos2B) - cos2C
( (= + 1 2

1 2A+2B 2A - 2B
2 2 2
= 2 - .2 cos ((, cos . - cos2C
((
= 2 - cos(A + B).cos (A - B) - cos2C

= 2 + cosC . cos(A - B) - cos2C

= 2 + cosC [cos(A - B)- cosC]

= 2 + cosC [cos(A - B) + cos (A + B)]

= 2 + 2cosA. cosB. cosC = RHS. Proved.

(b) LHS = cos2A + cos2B + cos2C

= 1 [2cos2A + 2cos2B] + cos2C
2
1
= 2 [1 + cos2A + 1 + cos2B] + cos2C

1 [2 + (cos2A + cos2B)] + cos2C
( (=2
1 2A+2B 2A-2B
=1+ 2 .2.cos 2 . cos 2 + cos2C

= 1 + cos(A +B) . cos(A -B) + cos2C

= 1 + (- cosC). cos (A -B) + cos2C

= 1 - cosC [cos(A -B)- cosC]

= 1 - cosC [cos(A -B) +cos(A +B)]

=1 - cosC . 2cosA . cosB = 1 - 2cosA. cosB. cosC =RHS. Proved.

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(c) LHS = sin2A - sin2B + sin2C

= 1 [2sin2A - 2sin2B] + sin2C = 1 [1–cos2A–1+cos2B] + sin2C
2 2
1
2 (cos2B - cos2A) + sin2C
( (=
1 2B+2A 2A-2B
2 2 2
= . 2 sin (( . sin + sin2C
(
= sin(A +B). sin(A -B) +sin2C
((
= sinC. sin(A -B) + sin2C
((
= sinC [sin (A - B) + sinC]((

= sinC [sin (A -B) + sin (A + B)]((

= sinC. 2sinA. cosB

= 2 sinA cosB.sinC = RHS. Proved.

Example 5. If A + B + C = Sc, then prove that
Solution:
(a) sin2 A + sin2 B - sin2 C = 1 - 2cos A . cos B sin C
238 2 2 2 2 2 2
A B C A B C
(b) cos2 2 + cos2 2 - cos2 2 = 2cos 2 . cos 2 sin 2

Here, A + B + C = Sc

or, A+B = Sc - C
2 2
( (sin Sc C
A+B = sin 2 - 2 = cos C
2 2
( (cos Sc C C
A+B = cos 2 - 2 = sin 2
2
Now,

(a) LHS = sin2 A + sin2 B - sin2 C
2 2 2
] ]=1 A B C
2 2 sin2 2 + 2sin2 2 - sin2 2

= 1 [1 - cosA + 1 - cosB] - sin2 C
2 2
1 C
=1 - 2 [cosA + cosB] - sin2 2
( (= 1
- 1 .2cos A+B . cos A-B - sin2 C
2 2 2 2
] ( ]= C A-B C
1 - sin 2 cos 2 + sin2 2
] ( ( ]=
1 - sin C cos A-B + cos A+B
2 2 2
C A B
= 1 - sin 2 . 2cos 2 . cos 2

= 1 - 2cos A . cos B . sin C = RHS. Proved.
2 2 2
A B C
2 + 2 - cos2 2
] ](b)LHS = cos2 cos2

= 1 2cos2 A + 2cos2 B - cos2 C
2 2 2 2

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vedanta Excel In Opt. Mathematics - Book 10

= 1 [1 + cosA + 1 + cosB] - 1 + sin2 C
2 2
1 C
= 1+ 2 [cosA + cosB] – 1 + sin2 2
( (=
1 A+B A-B C
2 2 2 2
. 2cos ((. cos + sin2
(= (
sin C . cos A-B ( + sin2 C
2 2 2
((
] ( ]= C
sin 2 . cos A-B + sin C
2 2
] ( ( ]=C
sin 2 . cos A-B + cos A-B
2 2
C A B
= sin 2 . 2cos 2 . cos 2

= 2cos A . cos B . sin C = RHS. Proved.
2 2 2

Example 6. (If A + B + C = Sc, prove thatABC Sc - A (cos Sc - B (. cos Sc - C
Solution: 2 2 2 4 4 4
cos + cos + cos = 4cos (( (
Example 7.
Solution: Here, A + B + C = Sc

or, A + B = Sc - C

B + C = Sc - A

C + A = Sc - B ? cos(C + A) = cos(Sc – B) = –cosB

Now,

LHS = cos A + cos B + cos C
2 2 2

= 2cos A/2 + B/2 . cos A/2 – B/2 + cos C + cosS2c
2 2 2
( ( ( (= 2 cos
A+B A-B C + Sc C - Sc
4 4 4 4
(( . cos + 2cos . cos((
( ( ( (= 2cos
Sc - C A-B Sc + C C - Sc
4 4 4 4
. cos(( + 2cos . cos((
( ] ( ( ]= 2cos
Sc - C A-B Sc + C
4 4 4
(( ( cos
+ cos ( cos(–T) = cos T
( ( (= 2cos
Sc - C A-B+Sc+ C A-B-Sc- C
4 8 8
(((
.2cos . cos
( ( (= 4cos
Sc - C Sc+A+C -B A-(B + C) - Sc
4 8 8
(( cos . cos (
( ] ( (= 4cos
Sc - C Sc+Sc-B -B A-Sc + A - Sc
4 8 8
((( cos . cos
( ( (= 4cos
Sc - C Sc - B A - Sc
4 4 4
( ((. cos . cos
( ( (= 4cos
Sc - A Sc - B Sc - C
4 4 4
.cos( (( cos = RHS. Proved.

If A + B + C = Sc, then prove that

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

We have A + B +C = Sc

or, A + B = Sc - C, B + C = Sc - A, C + A = Sc - B

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vedanta Excel In Opt. Mathematics - Book 10

Now,

LHS = cos (B + C - A) + cos (C + A - B) + cos (A + B - C)

= cos (Sc - A - A) + cos (Sc - B - B) + cos (Sc - C - C)

= cos (S - 2A) + cos (Sc - 2B) + cos(Sc - 2C)

= - cos2A - cos2B - cos2C

( (= - [cos2A + cos2B] - (2cos2C - 1)
2A+2B 2A-2B
2 2
= - 2cos (((. cos - 2cos2C + 1
((
= 1 - 2cos (A + B).cos(A -B)- 2cos2C

= 1 + 2cosC. cos(A -B) - 2cos2C

= 1 + 2cosC [cos(A - B) - cosC]

= 1 + 2cosC [cos(A - B) + cos (A +B)]

= 1 + 2cosC.2cosA. cosB

= 1 + 4 cosA. cosB. cosC = RHS. Proved.

Example 8. If A + B + C = 180°, then prove that
Solution: ( ( (sin(B+2C)+sin(C+2A)+sin(A+2B) = 4sin
B-C .sin C-A .sin A-B
2 2 2
Here, A + B + C = 180°

B + C = 180° - A

C + A = 180° - B , A + B = 180° - C

Now,

LHS = sin (B + 2C) + sin (C + 2A) + sin (A + 2B)

= sin (B + C + C) +sin (C + A + A) + sin (A + B + B)

= sin (Sc - A + C) + sin (Sc - B +A) + sin (Sc - C + B)

= sin{Sc - (A - C)} + sin {Sc -(B - A)} + sin {Sc - (C - B)}

= sin(A -C) +sin(B -A) + sin (C - B)
( (= 2sin
A-C+B-A A-C-B+A
2 2
(( . cos - sin (B - C)
( ( ( (= 2sin
B-C 2A - B - C B-C B-C
2 2 2 2
(( (. cos (
– 2sin . cos (
( ] ( ( ]= 2sin
B-C 2A - B - C B-C
2 2 2
(( (
cos – cos
( ] ( ( ]=2sin
B-C 2A - B – C + B – C B - C – 2A+ B + C
2 4 4
(( 2sin sin
( ( (= 2sin
B-C 2A - 2C 2B - 2A
2 4 4
(( (. sin . sin
( ( (= 4sin
B-C A-C B-A (
2 2 2
((. sin . sin ( sin(–T) = –sin T
( ( (= 4 sin
B-C C-A A-B
2 2 2
. sin((( . sin = RHS. Proved.

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Example 9. ( (If A + B + C = 180°, proved thatA A
Solution: tan 2 + tan B+C (= sec 2 .sec B+C (
2(( 2
Here, A + B + C = Sc, ((
( (sin
B+C (= sin Sc - C = cos C
2 2 2 2
( (and C
cos B+C = cos Sc - 2 = sin C
2 2 2
(LHS A B+C
= tan 2 + tan 2

= tan A + cotA2 = sinA/2 + cosA/2
2 cosA/2 sinA/2
sin2A/2 + cos2A/2
= sinA/2.cosA/2

= 1 = 1 . 1
sinA/2.cosA/2 sinA/2 cosA/2
(= cosecA2 . secA2
= sec A .sec B+C (= RHS. Proved.
2 2

Exercise 9.4

Very Short Questions

1. (a) Define conditional trigonometric identities with an example.

(b) If A,B and C are angles of a triangle ABC, then show that.

(i) sin(A + B) - sinC = 0

(ii) cos (B + C) + cosA = 0

(iii) tan(B + C) + tanA = 0

(c) If A + B + C = 180°, Show that :

(sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = sin (A - C) + sin (B - A) + sin(C - B) C
(d) If A, B, C are angles of a triangle, show that tan A+B ( = cot 2
2
Short Questions

2. If A + B + C = Sc, prove that

(a) tanA + tanB + tanC - tanA tanB.tanC = 0

(b) cotA.cotB + cotB. cotC + cotC . cotA - 1 = 0

(c) tan A ..ctoatnB2B2. + tan B . tan C + tan C . tan A =1
(d) cot A2 2 2 2 2
2 C = A + B C
cot 2 cot 2 cot 2 + cot 2

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

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Long Questions

3. If A + B + C = Sc, prove the following :

(a) sin2A + sin2B - sin2C = 4cosA.cosB.sinC

(b) sin2A - sin2B + sin2C = 4cosA.sinB.cosC

(c) cos2A + cos2B - cos2C = 1 - 4 sinA.sinB.cosC

4. If A + B + C = Sc, prove the following :

(a) sinA - sinB + sinC = 4sin A .cos B . sin C
2 2 2
A B C
(b) sinA + sinB - sinC = 4sin 2 .sin 2 . cos 2

(c) sinA - sinB - sinC = – 4cos A .sin B .sin C
2 2 2
A B C
(d) - sinA + sinB + sinC = 4cos 2 .sin 2 .sin 2

5. If A + B + C = 180° then prove that the following :

(a) cosA - cosB + cosC = 4cos A .sin B .cos C -1
2 2 2
A B C
(b) -cosA + cosB + cosC = -1 + 4 sin 2 .cos 2 .cos 2

(c) cosA - cosB - cosC = 1 - 4sin A .cos B . cos C
2 2 2
6. If A + B + C = Sc, prove the following :

(a) sin2A + sin2B - sin2C = 2sinA.sinB.cosC

(b) sin2A - sin2B - sin2C = - 2cosA.sinB.sinC

(c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC

7. If A + B + C = Sc, prove the following :

(a) cos2A - cos2B + cos2C = 1 - 2sinA.cosB. sinC

(b) cos2A - cos2B - cos2C = - 1 + 2cosA.sinB.sinC

(c) cos2A + cos2B - cos2C = 1 – 2sinA.sinB.cosC

8. If A + B + C = Sc, prove the following :

(a) sin2 A + sin2 B + sin2 C = 1 - 2sin A .sin B .sin C
2 2 2 2 2 2
A B C A B C
(b) sin2 2 - sin2 2 + sin2 2 = 1 - 2cos 2 .sin 2 .cos 2

(c) sin2 A - sin2 B - sin2 C = 2sin A .cos B .cos C - 1
2 2 2 2 2 2
9. If A + B + C = Sc, prove the following :

( )(a) A B C A B C
cos2 2 + cos2 2 + cos2 2 = 2 1 + sin 2 . sin 2 . sin 2

(b) cos2 A - cos2 B - cos2 C = - 2sin A . cos B .cos C
2 2 2 2 2 2
A B C A B C
(c) cos2 2 - cos2 2 + cos2 2 = 2cos 2 . sin 2 .cos 2

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10. If A + B + C = Sc, prove that

(a) sin A + sin B + sin C = 1 + 4sin Sc- A . sin Sc- B . sin Sc- C .
2 2 2 4 4 4
A B C Sc-A Sc-B Sc-C
(b) cos 2 + cos 2 + cos 2 = 4cos 4 . cos 4 . cos 4

(c) cosA + cosB + cosC = 1 + 4cos Sc-A cos Sc-B cos Sc-C
2 2 2
11. If A + B + C = Sc, prove that

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

12. If A + B + C = Sc, prove that

( ) ( ) ( )(a) cos(B+ 2C) + cos(C + 2A) + cos(A + 2B) = 1 - 4cos
A -B .cos B-C .cos C -A
2 2 2
( ) ( ) ( )(b) sin(B+ 2C) + sin(C + 2A) + sin(A + 2B) = 4sin A -B C -A
2 .sin B-C .sin 2
2
(c) sin3A + sin3B + sin3C = – cos 32A.cos32B.cos32C
(d) cos4A + cos4B +cos4C = – 1 + 4cos2A . cos2B . cos23C

13. If D + E + J = Sc , prove that
2
(a) sin2D + sin2E + sin2J = 1 - 2 sinD. sinE.sinJ

(b) tanE.tanJ. + tanJ . tanD + tanD . tanE = 1

(c) cos(D - E - J) + cos (E - J - D) + cos(J - D - E) = 4cosD.cosE.cosJ

14. If A + B + C = Sc, prove that

(a) sin2A + sin2B + sin2C = 8 sinA2 sinB2 sin C
4cosA2 .cosB2 .cosC2 2

(b) cosA + cosB + cosC = 2
sinB.sinC sinC.sinA sinA.sinB

(c) sinA + sinB + sinC = 2tanA. tanB.tanC
cosB.cosC cosC.cosA cosA.cosB

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

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

15. If A + B + C = Sc, prove that

tanA + tanB + tanC + tanB + tanC + tanA
tanB tanC tanA tanA tanB tanC

= cosA.secB.secC + cosB.secC.secA + cosC.secA.secB

16. If A + B + C = 2S, prove that
sin(S – A).sin(S – B) + sinS.sin(S – C) = sinA.sinB

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vedanta Excel In Opt. Mathematics - Book 10

9.5 Trigonometric Equations

Let us consider the following equalities:

(a) sin2T + cos2T = 1 (b) 4sin2T = 1

(c) sec2T = 1 + tan2T (d) tanT = 1
3

Discuss the following questions from above statements.

(a) Which are identities ?

(b) Which are equations ?

(c) What are differences between identities and equations ?

Equalities like sin2T + cos2T = 1, sec2T = 1 + tan2T are satisfied by all values of the
angle T. So they are called identities.

But 4sin2T = 1 and tanT = 1 etc. are satisfied by only some values of the angle T. So they
3
are called trigonometric equations.

Definition: An equation containing the trigonometrical ratios of an unknown angle is called
trigonometric equation.

Example : sinT = 3 , 0° ≤ T ≤ 360°
2
For 0° ≤ T ≤ 360°,

sinT = 3
2
or, sinT = sin60° or sin120°

? T = 60° or 120°

To solve such trigonometric equations, we need some limitations for the value of variable.
This limitation is called the range of the variables.

We note that the values of sinT and cosT repeat after an interval Sc. If the equation involves a
variable T, 0 ≤ T ≤ 2Sc, then the solutions are called principal solutions. A general solution
is one which involves the integer 'n' and gives all solutions of trigonometric equation. But
we study only about principal solutions.

Example : Find the solution of cosT = 1 , 0° ≤ T ≤ 360°
2
We know that

cosT = cos(360° - T) = cosT

or cosT = 1 = cos30°
2
? T = 30°

or, cosT = 1 = cos(360° - 30°) = cos30° or cos330°
2

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Note :
Let T be any angle. Then we note the following :

i. The value of sinT lies between -1 and +1. Y
i.e. -1 ≤ sinT ≤ 1, otherwise there is no

solution. (180° – T),S (360° + T),

ii. The values of cosT lies between -1 and (sin and cosec positive) (All positive)
X' X
+1. i.e. -1≤ cosT ≤ 1, otherwise there is TO C
no solution. (tan and cot positive) ( cos and sec positive)

iii. The value of tanT is infinitely positive or (180° + T) (360° – T)

infinitely negative. it has no boundary of Y'
its values. i.e. - ∞ < tanT < + ∞

iv. Find the quadrants where the sign of value of trigonometric ratio of angle falls by
using the rule CAST.

Some Basic Ideas

i. If sinT = sinA, then T = A or 180° - A

Example : Solve : sinT = 1
2
1
sinT = 2 = sin45°

Since sine is positive in the first and the second quadrants, we write,

sinT = sin45° or sin(180° - 45°)

? T = 45° or 135°

ii. If sinT = - sinA, then T = 180° + A or, 360° - A

Example : Solve sinT = - 1
2
sinT = - sin45°

Since sine is negative in the third and the fourth quadrant we write,

sinT = sin (180° + 45°) or sin(360° - 45°)

? T = 225° or 315°

iii. If cosT = cosA, thenT = A or (360° - A)

Example : Solve cosT = 3
2
? cosT = cos30°

Since cos is positive in the first and the fourth quadrant, we write,

cosT = cos30° or cos(360° - 30°)

? T = 30° or, 330°

iv. If cosT = -cosA, then T = 180° - A or, 180° + A

Example : Solve : cosT = - 3 , cosT = - cos30°
2

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Since cos is negative in the second quadrant and third quadrant, we write,

cosT = cos (180° - 30°) or cos (180° + 30°)

? T = 150° or 210°

Working steps for solution of Trigonometric equations

The following are some hints that will be helpful in solving a trigonometric equations :

(i) Express all the trigonometric functions in terms of a single trigonometric function of
same angle if possible.

(ii) Transfer every terms of the equation to the left hand side.

(iii) If equation is quadratic, factorize the left hand side and equate each factor to zero.

(iv) If the equation is quadratic in certain trigonometric function use formulae for the
solution (in case not factorizable).

Note :
In the process of solving trigonometric equations if there is involvement of squaring
or cubing, which give rise to some additional equations and consequently some
additional roots. All the solutions thus obtained may not satisfy the given equation.
The values which do not satisfy are discarded: only the values which satisfy be given
equations are accepted. Each is to be checked. Only accepted values are the solution
of the given equation.

Example : sinT - cosT = 1 0 ≤ T ≤ 360°
Solution:
We can write, sinT = 1 + cosT
246
Squaring on both sides, we get

sin2T = 1 + 2cosT + cos2T

or, cos2T + 2cosT + 1 - 1 + cos2T = 0

or, 2cos2T + 2cosT = 0

or, 2cosT (cosT + 1) = 0

Either, 2cosT = 0 ................ (i)

cosT + 1 = 0 ........... (ii)

From equation (i), cosT = 0

or, cosT = cos90°, cos270°, ? T = 90°, 270°

or, cosT = -1 =cos180°

? T =180°

Hence, T = 90°, 180°, 270°

on checking, put T = 90°, in given equation

sin90° - cos90° = 1

sin90° - cos90° = 1

or, 1 = 1 (It is true.)

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vedanta Excel In Opt. Mathematics - Book 10

Putting T = 180°,
sin180° - cos180° = 1

or, 0 + 1 = 1
? 1 = 1 (true)
Putting T = 270.

sin270° - cos270° = 1
or, -1 -0 = 1
or, -1 = 1 (false)
Hence, 270° is rejected.
? T = 90°, 180°

Some special cases for 0° ≤ T ≤ 360°
(i) If sinT = 0, then T= 0°, 180°, 360°
(ii) If cosT= 0, thenT = 90°, 270°
(iii) If tanT = 0, thenT = 0, 180°, 360°
(iv) If sinT = 1, then T = 90°
(v) if sinT = -1, then T = 270°
(vi) If cosT = -1, then T = 180°
(vii) If tanT = ∞, then T = 90°, 270°
(viii)90° × n ± , when n = odd number (i.e. 1, 3, 5, .....)

sin œ cos, tan œ cot, sec œ cosec
when n = even number (i.e. 2, 4, 6, .............)
sin œ sin, cos œ cos, tan œ tan

Worked Out Examples

Example 1. Solve the equation: (0° ≤ T ≤ 180°)
Solution:
(a) sinT = 3 , (b) tanT = 3 (c) cosT = 1
(a) Here, 2 = 2
3
sinT 2

since, 0 ≤ T ≤ 180°

sinT = 3 = sin60° or sin(180° - 60°)
2
? T = 60° Ÿ 120°

(b) Here, tanT = 3

since 0° ≤ T ≤ 180°

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vedanta Excel In Opt. Mathematics - Book 10

tanT = 3 = tan60°

? T = 60°

(c) Here, cosT = 1
2

since, 0° ≤ T ≤ 180°

cosT = 1 = cos45°
2

? T = 45°

Example 2. Solve the equation: 0° ≤ T ≤ 360°.
Solutions:
(a) 3 tanT = 1, (b) cosec2T = 2 (c) 3tan2T - 1 = 0

(a) Here, 3 tanT = 1

or, tanT = 1
3

Since tanT is positive, T lies in the first or the third quadrant.

? tanT = tan30° or, tan (180° + 30°)

? T = 30° or, 210°

(b) cosec2T = 2

or, 1 =2 or, sin2T = 1
sin2T 2
1
? sinT = ± 2

Taking positive sign, we get

sinT = 1
2

Since, sinT is positive, T lies in the first or second quadrant,

sinT = sin45° or, sin(180° - 45°)

? T = 45° or, 135°

Taking negative sign, we get

sinT = - 1
2

Since sinT is negative it lies in the third or fourth quadrant.

? sinT = sin(180° + 45°) or, sin(360° - 45)

? T = 225° or, 315°

Hence, the required values of T are 45°, 135°, 225° or 315°

(c) Here, 3 tan2T - 1 = 0

( )or, 1 2
tan2T = 3

or, tanT = ± 1
3

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Taking positive sign,

tanT = 1
3

Since tanT is positive, T lies in the first or third quadrant

? tanT = tan30° or, tan (180° + 30°)

? T = 30° or 210°

Taking negative sign, we get

tanT = - 1
3

Since tanT is negative, T lies in the second or fourth quadrant.

? tanT = tan(180° - 30°) or, tan (360° - 30°)

? T = 150° or, 330°

? T = 30°, 150°, 210°, 330°.

Example 3. Solve : tan 9T = cotT, 0° ≤ T ≤ 90°
Solution:
Here, tan 9T = cotT

or, tan9T = tan(90° - T), [cotT = tan(90° - T)]

tan(270° - T), tan(7×90° - T), tan(5×90° - T), tan(9×90° - T)

? 9T = 90° - T, 270° - T, 450° - T, 630° - T, 810° - T

or, 10T = 90°, 270°, 450°, 630°, 810°

? T = 9°, 27°, 45°, 63°, 81°

Example 4. 4sin2T + 3cosT = - 3, 0° ≤ T ≤ 360°
Solution:
Here, 4sin2T + 3cosT = - 3

or, 4 - 4cos2T + 3cosT + 3 = 0

or, - 4cos2T + 3cosT + 7 = 0

or, 4cos2T – 3cosT – 7 = 0

or, 4cos2T - 7cosT + 4cosT - 7 = 0

cosT (4cosT - 7) + 1 (4cosT - 7) = 0

or, (4cosT - 7) (cosT + 1) = 0

Either, 4cosT - 7 = 0 ........... (i)

cosT + 1 = 0 ............... (ii)

From equation (i) cosT = 7 >1
4
since, - 1 ≤ cosT ≤ 1, it has no solution.

Hence, cosT = 7 is rejected.
4
From equation (i), we get,

cosT = - 1

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or, cosT = cos180°

? T = 180°

Hence, the required solution is T = 180°.

Example 5. 2cosT = 3 cotT, 0° ≤ T ≤ 360°
Solution:
Here, 2cosT = 3 cotT
Example 6.
Solution: or, 2cosT = 3 cosT (Note: Do not cancel cosT)
sinT

or, 2cosT.sinT - 3 cosT = 0

or, cosT (2sinT - 3 ) = 0

Either, cosT = 0 .............(i)

or, 2sinT - 3 = 0 .............. (ii)

From equation (i), cosT = 0

or, cosT = cos90°, cos270° ? T = 90°, 270°

From equation (ii), sinT = 3
2
or, sinT = sin60° or, sin (180° - 60°)

? T = 60° or, 120°

Hence, the required solutions are x = 60°, 90°, 120°, 270°

Solve : 3 sinT - cosT = 1, 0° ≤ T ≤ 360°

Here, 3 sinT - cosT = 1............ (i)

coefficient of sinT = 3

coefficient of cosT = - 1

( 3)2 + (-1)2 = 3 + 1 = 2

Dividing equation (i) by '2' on both sides, we get,

3 sinT - 1 cosT = 1
2 2 2
1
or, cos30° . sinT - sin30° cosT = 2

or, sin(T - 30°) = sin30° or, sin (180° - 30°)

? T - 30° = 30° or, 150°

or, T = 30° + 30° or, 150° + 30°

? T = 60°, 180° are required solutions.

Alternative Method

Here, 3 sinT = 1 + cosT

Squaring on both sides, we get,

3sin2T = 1 + 2cosT + cos2T

or, 3(1 - cos2T) = 1 + 2cosT + cos2T

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or, 2cosT + cos2T + 1 = 3 - 3cos2T

or, 4cos2T + 2cosT - 2 = 0

or, 2cos2T + cosT - 1 = 0

or, 2cos2T + 2cosT - cosT - 1 = 0

or, 2cosT (cosT + 1) - 1 (cosT + 1) = 0

or, (2cosT - 1) (cosT + 1) = 0

Either, 2cosT - 1 = 0 ........ (i)

cosT + 1 = 0 .......(i)

From equation (i), cosT = 1 ,
2
or, cosT = cos60° or, cos(360° - 60°)
T = 180°
? T = 60° or 300°
T = 300° is rejected
From equation (ii) cosT = -1,

or, cosT = cos180° ?

? T = 60°, 180°, 300°

On checking,

For, T = 60°

3 sinT - cosT = 1

or, 3 . sin60° - cos60° = 1

or, 3 . -2312- 1 = 1
or, 3 2
2
=1

or, 1 = 1 (True)

For T q
or, 3 sin180° - cos180° = 1

or, 3 . 0 - (-1) = 1

? 1 = 1 (True)

For T = 300°

or, 3 sin300° - cos300° = 1

( )or, 3 - 3 - 1 =1
2 2
3 1
or, - 2 - 2 =1

or, -2 = 1 (False) ?

? T = 60°, 180° are required values.

Example 7. Solve cos3T + cosT = cos2T, 0° ≤ T ≤ 180°
Solution: Here, cos3T + cosT = cos2T

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( ) ( )or,
2cos 3T + T . cos 3T - T = cos2T
2 2
or, 2cos2T.cosT - cos2T = 0

or, cos2T (2cosT - 1) = 0

Either, cos2T = 0 ........ (i)

2cosT - 1 = 0 ........ (ii)

From equation (i), cos2T = 0

or, cos2T = cos90°, cos270°

? 2T = 90°, 270°

? T = 45°, 135°

From equation (ii),

2cosT - 1 = 0

or, cosT = 1
2
or, cosT = cos60°

? T = 60°, 300°

? T 45°, 60°, 135° are the required values.

Example 8. Solve 3 + 1 = 4 , (0° ≤ T ≤ 90°)
Solution: sin2T cos2T

Example 9. Here, 3 + 1 = 4
Solution: sin2T cos2T

or, 3 cos2T + sin2T = 4cos2T.sin2T

Dividing both sides by 2, we get,

3 cos2T + 1 sin2T = 2sin2T.cos2T
2 2
or, sin60°. cos2T + cos60°. sin2T = sin4T

or, sin(2T + 60°) = sin4T

or, sin4T = sin(2T + 60°)

or, 4T = 2T + 60°

or, 2T = 60°

? T= 30°
Solve : tanT + tan2T + tanT . tan2T = 1, 0° ≤ T ≤ 360°

Here, tanT + tan2T + tanT.tan2T = 1

or, tanT + tan2T = 1 - tanT. tan2T

or, tanT + tan2T = 1
1 - tanT . tan2T

or, tan 3T= tan45°, tan (180° + 45°), tan(360° + 45°), tan(540° +45°),

tan(720° + 45°), tan(900° + 45°)

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? 3T = 45°, 225°, 405°, 585°, 765°, 945°
? T = 15°, 75°, 135°, 195°, 255°, 315°

Example 10. Solve : tan2T - (1 + 3 ) tanT + 3 = 0, (0° ≤ T ≤ 360°)

Solution: Here, tan2T - (1 + 3 ) tanT + 3 = 0

or, tan2T - tanT - 3 tanT + 3 = 0

or, tanT (tanT - 1) - 3 (tanT - 1) = 0

or, (tanT - 1) (tanT - 3) = 0

Either, tanT - 1 = 0 ....... (i)

tanT - 3 = 0 ........ (ii)

From equation (i), tanT = 1

or, tanT = tan45°, tan225°

From equation (ii), tanT = 3

or, tanT = tan60°, tan240°

? T = 60°, 240°

? T = 45°, 60°, 225°, 240° are the required values.

Example 11. If tanT + tanE = 2 and cosv.cosE = 1 solve for values of D and E
2
Solution: Here, tanD + tanE = 2

or, sinD + sinE =2
cosE cosD

or, sinD.cosE + cosD.sinE = 2cosD. cosE

or, sin(D + E) = 2cosD. cosE ....... (i)

But cosD.cosE = 1 ......... (ii)
2
1
? sin(D + E) = 2 .2

or, sin(D + E) = 1 = sin90°

? D + E = 90°

and E = 90° - D

From (ii) 2cosD.cos(90° - D) = 1

or, 2cosD. sinD = 1

or, sin2D = sin90°

or, 2D = 90° ? D = 45°

Also, D + E = 90°

Ÿ E = 90° - D = 90° - 45° = 45°

? D = 45°, E = 45°

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Exercise 9.5

Very Short Questions

1. Find the value of T in the following equation, 0° ≤ T ≤ 180°.

(a) 2sinT = 3 (b) 3 tanT = 1 (c) cosT = 1
(d) cotT = 3 (e) secT = 2 2
(f) 2cosT + 1 = 0

(g) 3 tanT - 1 = 0

2. Find the value of T, 0° ≤ T ≤ 90°

(a) tanT = cotT (b) cosT = sinT (c) secT = cosecT

(d) cot2T = tanT (e) sin4T = cos2T (f) cos2T = sin3T

3. Solve the following questions : 0° ≤ T ≤ 180°

(a) sin2T = sinT (b) tanT = sinT (c) 3cotT - tanT = 2

(d) 2 +2=0 (e) tanT + cotT = 2 (f) 1 + cos2T = cosT
cosT
Short Questions

4. Solve, 0° ≤ T ≤ 360° (b) 4sin2T = 3 (c) sec2T = 4
(a) 3tan2T = 1 (e) 3sec2T - 4 = 0
(d) sec2T = 2tan2T

Long Questions

5. Solve for T, 0° ≤ T ≤ 180°

(a) sin4T = cosT - cos7T (b) sinT + sin2T + sin3T = 0
(d) cos2T + cos4T = cos3T
(c) sinT - sin2T = 0 (f) 3cotT - tanT = 2

(e) secT . tanT = 2

6. Solve : 0° ≤ T ≤ 360° (b) 2sin2T - 3sinT + 1 = 0
(a) 2cos2T - 3sinT
(c) 2cos2T - 5cosT + 2 = 0 (d) 2sin2T + sinT - 1 = 0
(e) 4cos2T + 4sinT = 5
(f) 3 - 2sin2T = 3cosT
( )(h) cot2T +
(g) tan2T + (1 - 3 ) tanT = 3 3 1 cotT + 1 = 0
7. Solve : 0° ≤ x ≤ 360° 3

(a) 3 cosx + sinx = 3 (b) sinx + cosx = 2
(d) 3 cosx = 3 - sinx
(c) sinx + cosx = 1, (f) cosx - 3 sinx = 1

(e) 3 sinx + cosx = 2

(g) 2 secx + tanx = 1

8. Solve for x, 0° ≤ x ≤ 180°

(a) 3 + 1 =0 (b) 3 + 1 = 4
sin2x cos2x sin2x cos2x

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9. Solve : 0° ≤ x ≤ 360°
(a) 2 tan3x.cos2x + 1 = tan3x + 2cos2x (b) sin2x . tanx + 1 = sin2x + tanx

10. Find the value of T from.
tanT - 3cotT = 2tan3T, 0° ≤ T ≤ 360°

11. Solve: 0° ≤ x ≤ 90°
tan(x + 100°) = tanx.tan(x + 50°).tan(x – 50°)

1. (a) 60°, 120° (b) 30° (c) 60° (d) 30°

(e) 45° (f) 120° (g) 30°

2. (a) 45° (b) 45° (c) 45° (d) 30°

(e) 15°, 75° (f) 18°, 90°

3. (a) 0°, 60°, 180° (b) 0°, 180° (c) 45°, 108.43° (d) 135°

(e) 45° (f) 60°, 90°

4. (a) 30°, 150°, 210°, 330° (b) 60°, 120°, 240°, 300°

(c) 60°, 120°, 240°, 300° (d) 45°, 135°, 225°, 315°

(e) 30°, 150°, 210°, 330°

5. (a) 0°, 10°, 45°, 50°, 90°, 130°, 135°, 170° (b) 0°, 90°, 120°, 180° (c) 0°, 60°, 180°

(d) 30°, 60°, 90°, 150°, 180° (e) 45°, 135° (f) 45°, 108.43°

6. (a) 30°, 150° (b) 0°, 30°, 90° (c) 60°, 300° (d) 30°, 150°, 270°

(e) 30°, 150° (f) 0°, 60°, 300°, 360°

(g) 60°, 135°, 240°, 315° (h) 120°, 150°, 300°, 330°

7. (a) 0°, 60°, 360° (b) 45° (c) 0°, 90°, 360° (d) 0°, 60°, 360°

(e) 60° (f) 0°, 240° (g) 315°

8. (a) 60°, 150° (b) 20°, 30°, 50°, 80°, 120°, 140°, 170°

9. (a) 15°, 30°, 75°, 150°, 210°, 225°, 315°, 330° (b) 45°, 225°

10. 45°, 60°, 120°, 135°, 225°, 240°, 300°, 315° 11. 30°, 55°

9.6 Height and distance

Trigonometric knowledge can be used to calculate heights of all objects or distance of

inaccessible objects. A

Introduction :

Let AB be a pillar. An observer is at point C which is at a distance BC from 12cm
the foot of the pillar. The observer observes the top of the pillar through
CA. The line CA is called sight line. AB is the height of the pillar. BC is the B 60°
distance between the foot of the pillar and the observer. Let AC = 12m and C
‘BCA = 60°. Now discuss the following questions.

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(i) What is the height of the pillar ?
(ii) Find the distance between the foot of the pillar and the observer.

Let us define some basic terms used in this topic : Object A

Angle of elevation :

When an object is at a higher level than that of the Sight line
observer, the observer has to have a view of the object Angle of elevation
or the observer has to look up. The angle which the

sight line makes with the horizontal the observers Observe's eye O T
eye is known as the angle of elevation. In the figure
‘AOB = T is the angle of elevation. O Horizontal Line through B
T observe's eye A
Angle of Depression :

When the object is at a lower level than that of the observer Angle of Depression Sight line
then he has to look below in order to see the object. The C B
angle which the sight line through the observer's eye is
known as the angle of depression. In the above figure

‘AOB = T is the angle of depression. OA is parallel to

CB, ‘AOB = ‘CBO = T, being the corresponding angles.

How is the angle of elevation or depression measured?
Special instruments like theodolite, sextant, clinometer,
hypsometer etc are used to measure the angle of depression.

Step 1 : Draw the appropriate figure which represents the
given condition.

Step 2 : Identify the given sides and angles in the figure.

Step 3 : Choose the appropriate trigonometric ratio to find the unknown sides or angles.

Step 4 : Carry out the necessary calculations with appropriate units.

Worked Out Examples

Example 1. Find the values of x and y from the given figures:

(a) A (b) S 45° P

xx

D 30° 60° B Q 40m R
20m Cy

Solution: (a) In the given figure from right angle ∆ABC,

tan60° = x
y

or, 3 y = x ? x = 3 y ............ (i)

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Again, from right angled ∆ABD,

tan30° = 20 x y
+

or, 1 = x y
3 20 +

or, 20 + y = 3 x .......... (ii)
Putting the value of x in equation (i), we get

20 + y = 3. 3y
or, 20 + y = 3y

or, 2y = 20 ? y= 20 ? y = 10m
2

Putting the value of y in equation (i)

20 + 10 = 3 x

or, x= 30 = 10 3 ? x = 10 3m
3

(b) In the given figure, SP//QR, corresponding angles.

‘SPQ = ‘PQR,

Now, from right angled triangle PQR,

tan45° = x or, 1 = x
40 40

? x = 40m

Example 2. A man observes the angle of elevation of the top of a tower to be 30°. On
Solution: walking 200 metre nearer, the elevation is found to be 60°. Find the height of
the tower. Also, find the distance between the first point from the foot of the
tower.

Let CD be the height of the tower, A and B be the first and the second
point respectively. In the figures AB = 200m, suppose BC = ym and

CD = xm. From right angled triangle BCD,

tan60° = CD
BC

or, 3 = x
y
D

or, x = 3 y .......... (i)

Again, from right angled triangle ACD. x

tan30° = CD 30° 60°
AC 200m By
A C
x
or, 1 = 200 + y
3

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or, 200 + y = 3 x ........... (ii)
Putting the value of x in equation (ii)

200 + y = 3. 3 y
or, 200 + y = 3y

or, 200 = 2y

? y = 100m

Putting the value of y in equation (i)

x = 3. 100 = 1.732 × 100 = 173.2 m
? The height of the tower is 173.2m and distance between the tower and

first point is (200 + 100) = 300m

Example 3. From the roof of a house, an observer finds that the angles of depression of
Solution: two places in the same side are 60° and 45°. If the distance between these
places is 40m, find the distance of the places from the foot of the house.

Let PQ be the height of the house and R and S he two points on the same side
of the house.

Then PT//QS, ‘TPR = ‘PRQ = 60°, ‘TPS = ‘PSR = 45°, RS = 40m.

Let QR = y m and PQ = x m. P 60°45° T

Now, from right angled triangle PRQ

tan60° = PQ xm
QR
x
or, 3 = y Q 60° 45° S
y R 40m
or, x = 3 y ............... (i)

Again, from right angled triangle PQS

tan45° = PQ
QS
x
or, 1 = y + 40

or, x = y + 40 ........ (i)

Putting the value of x in equation (ii) from (i), we get

3 y = y + 40

or, y ( 3 - 1) = 40

or, y = 5434.06-14m= 40 - 1 = 40
? y = 1.732 0.732

Putting the value of y in equation (i)

x = 3 × 54.64 = 94.64m

? The height of the house (x) = 94.64m

Distance of the first point from the house (y) = 54.64m

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Distance of the second point from the house (x) = y + 40 = 54.64 + 40
= 94.64m.

Example 4. Two girls are on the opposite sides of a tower which is 200m high. They
Solution: observed that the angles of elevation of the top of the tower are 30° and 45°
respectively. Find the distance between them.

Let A and B be the positions of the two girls on opposite sides of the tower DC.

DC = 200m.

The angles of elevations are

‘DAC = 30°, ‘DBC = 45°

Let AC = xm and BC = ym

From right angled ∆DAC, D

tan30° = DC 200m
AC
1 200
or, x =3 = x A 30° 45° B
or, 200 3 .......(i) xm C ym

Again, from right angled ∆ACB,

tan45° = CD
CB
200
or, 1= y ? y = 200m
= AB = AC + CB
Distance between the girls

= x + y = 200 3 m + 200m = 546.41m

Example 5. The angle of depression of the top and the foot of a lamp post observed from
Solution: the roof of a 60m high house and found to be 30° and 60°. Find the height of
the lamp post.

Let MN and PQ be the height of the house and lamp post respectively.

Then, MN = 60m, ‘RMP = ‘MPS = 30°, ‘RMQ = ‘MQN = 60°

From right angled, ‘MQN, R 30° M
60°
MN
tan60° = QN

or, 3 = 60 P 30° S
QN
60 60
? QN = 3 = 3 3 = 20 3 m. Q 60°

N

PS = QN = 20 3 (opposite sides of a rectangle)

Again, from right angled triangle MPS,

tan30° = MS
PS
1 MS
or, 3 = 20 3

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MS = 20 3 = 20m
3

Hence, the height of the lamp post is

PQ = SN = MN - MS = 60 - 20 =40m.

Example 6. A boy standing between two pillars of equal height observes the angle of
Solution: elevation of the top of pillars to be 30°. Approaching 15m towards any one
of the pillar, the angle of elevation is 45°. Find the height of the pillars and
the distance between them.

Let PQ and MN be two equal pillars of equal height,

QN the distance between the pillars.

O is mid point of QN. MN = PQ.

‘MON = 30°, ‘POQ = 30°, RO = 15m.

From right angled triangle PQR, tan45° = PQ
QR
or, QR = PQ

Again, from right angled triangle PQO,

tan30° = PQ P M
QO N
1 PQ
or, 3 = QR + 15

1 PQ Q 45° 30° 30°
3 PQ + 15 R 15m O
or, =

or, 3 PQ = PQ + 15

or, ( 3 - 1 ) PQ = 15

or, PQ = 15 = 15 × 3 +1
3-1 3-1 3 +1
15 ( 3 +1)
= 3-1

= 15 (1.732 + 1) = 7.5 × 2.732
2

= 20.49m.

Again, from the right angled triangle MON,

tan30° = MN or, 1 = 20.49
ON 3 ON
? ON = 35.49m

Distance between two pillars is 2 × ON = 2 × 35.49 = 70.98m.
Hence, the height of the pillar = PQ = MN = 20.49m

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Example 7. A flagstaff stands upon the top of a building. Find the length of the flagstaff
Solution: and building observed from a point at a distance of 20 m from the building
are respectively 60° and 45°.

Let PQ be the flagstaff and QR be the building respectively. P

Let S be a point 20m away from the building, SR = 20m

Now, from right angled triangle SRQ. Q

tan45° = QR 60°
SR 45°
S R
20m

or, 1= QR
20
? QR = 20m

Again, from right angled triangle PRS.

tan60° = PR
SR

or, 3 = PQ + QR
SR

or, 3= PQ + 20
20

or, 20 3 = PQ + 20

? PQ = 20 ( 3 - 1) = 20 (1.732 - 1)

= 20 × 0.732 = 14.64m.
Hence, the height of the flagstaff is 14.64m.

Example 8. PQ and MN are two towers standing on the same horizontal plane. From
Solution: point M the angle of depression of Q is 30° and from N the angle of elevation
of P is 60°. If the shorter tower is MN = 100m, find the height of the tower
PQ.

In given two towers are PQ and MN, MN<PQ. MN = 100m,

‘RMQ = ‘MQN = 30°

‘PNQ = 60°. P
From right angled triangle MQN,

tan30° = MN
QN

or, 1 = 100 R 30° M
3 QN Q 30° 60° N
100m
? QN = 100 3 m. 261

RMNQ is a rectangle,

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? RM = QN = 100 3 m.
From right angled triangle PQN,

tan60° = PQ
QN
PQ
or, 3= 100 3

? PQ = 300m

? The height of the longer tower PQ is 300m.

Example 9. From a light house the angles of depression of two ships on the opposite
Solution: sides of the light house were observed to be 30° and 45°. If the height of the
light house is 200m and the line joining the two ships passes through the
foot of the light house, find the distance between the ships.

Let AD be the height of the light house.

‘EAB = ‘ABD = 45°, ‘FAC = ‘ACD = 30°

From the right angled triangle ABD, tan45° = AD
BD
or, BD = AD
A
? BD = 100m E F
100m
Again, from right angled triangle ADC,

tan30° = AD B 45° D 30° C
DC

or, 1 = 100
3 DC
or, DC = 173.2m.

Hence, the distance between the ships is given by

BC = BD + DC = 100m + 173.2m = 273.2m

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

Solution: Let MN be the height of the flagstaff and PS be a ladder.

Then ‘MPN = 60°, MS = 9m, PS = 9m M
Let SN = x, PN = y .

From right angled triangle MPN, by using Pythagoras theorem, 9m

PS2 = PN2 + SN2 P 60°y S
or, 81 = y2 + x2 x
? x = 81 - y2 ......... (i)
N

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Again, from right angled triangle MPN,

tan60° = MN
PN

or, 3= 9+x
y

or, 3 y = 9 + x ............ (ii)

By solving equation (i) and (ii), we get

3 y – 9 = 81 – y2

squaring on both sides, we get,

( 3 y – 9)2 = ( 81 – y2)2

or, 3y2 – 18 3y + 81 = 81 – y2

or, 3y2 + y2 = 18 3y

or, 4y2 = 18 3y

or, 4y2 = 18 3
y
18 3
or, y = 4

since y ≠ 0, y = 93 = 7.8
2

and x = 81 – 60.75 = 4.5m

? MN = 9 + 4.5 = 13.5m

Alternative Method

In above diagram, ∆PSM is an isosceles triangle.

? ‘PMS = ‘MPS = 30°

‘SPN = 60° - 30° = 30°

From, right angled triangle ∆SPN

sinSPN = SN
PS
x
or, sin60° = 9

or, 1 = x
2 9

? x = 4.5 m

Hence, the height of the flagstaff is (9 + 4.5) = 13.5m

Example 11. A vertical column is divided into ratio 1:9 from the bottom by a point on
it. If these two parts make an equal angle 20m from the bottom of the any
column, find the height of the column.

Solution: Let MN be the height of the vertical column and P be the point on it which
divides MN in the ratio of 1:9 from the bottom.

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Let NP = x and PM = 9x.

Let each segment of the column makes an angle T at Q.

‘PQN = T ‘MQP = T, QN = 20m M

From right angled triangle PQN 9x

tanT = PN = x P
QN 20 x
Again, from right angled triangle MQN,
N
tan2T = MN Q T
QN
T
20m

or, 2tanT = 10x
or, 1- -ta2nx02T 20
1
= x
x2 2
1 - 400

or, x × 400 = x
10 400 - x2 2
or, 400 - x2 = 80

or, x2 = 320

Since height is positive, taking positive square root only,

x = 8 5 m.

The height of the column is 10x = 10 × 8 5 = 80 5 m.

Example 12. The angle of elevation of a bird from a point 80m above the lake is 30°.
From the same point, the angle of depression to its image on the same lake
is found to 60°. Find the height at which the bird is at the moment.

Solution: Let B be the position of the bird and its image is B' in the lake. Then a point A
is 80m above the surface of water, XY the level of water of the lake.

Let BC = x, CB' = y, AD = 80m. ‘CAB = 30°, ‘B'AC = 60°

From right angle triangle ABC, B
x
tan30° = BC C 30°
CA
1 x 60° A
or, 3 = CA XE 80m
DY
or, CA = 3 x ........(i) B'

Again, from right angled triangle ACB'

or, tan60° =3y.xCCBA'[by using (i)]
3=

or, y = 3x ............ (i)

But y = CE + EB' ( BE = B'E)

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B'E = 80 + BC = 80 + x

? y = 80 + 80 + x ....... (iii)

From (ii) and (iii), we get,

160 + x = 3x

or, 2x = 160 ? x = 80m

? The position of the bird from the level of lake is

BE = CE + CB = 80 + 80 = 160m

Example 13. The shadow of a pole on the ground level increases in length by x metres
when the sun's attitude changes from 45° to 30°. Calculate values of x given
that height of the pole is 50m.

Solution: Let MN be the height of the pole, MN = 50m.

Let NP = y and NQ = y + x be the length of shadows when sun's altitudes are
45° and 30° respectively. Increased in length of shadow is PQ = x.

From, right angled triangle MPN, M
50m
tan45° = MN
PN
50
or, 1 = y

or, y = 50m Q 30° 45° N
Again, from right angled triangle MQN. xPy

tan30° = MN
QN

or, 1 = 50
3 x+y

or, 50 3 = x + 50
or, x = 50 ( 3 - 1)

or, x = 50 × (1.732 - 1) = 50 × 0.732 = 36.6m

? x = 36.6m

Exercise 9.6

Very Short Questions

1. Define the following terms with figures:

(a) Angle of elevation (b) Angle of depression

2. Find the value of x in the following diagrams:

PM 30° N
20m
(a) (b)

60m x

Q 60° R P xO

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Short Questions

3. Find the values of x and y in the given diagrams:

P

(a) (b) P

y 45°

S 30° 45° Q 200m
20m R x
Q 60° R
x Sy

(c) P (d) P 30° 60° U

Q 9m 9m 20m 30° R
60° S T x 60° S
x y
y R
Q

Long Questions

4. (a) The angle of elevation of the top of a house from a point on the ground was
observed to be 60°. On walking 60m away from that point, it was found to be 30°.
If the house and points are in the same line of the same plane, find the height of
the house.

(b) The angle of elevation of the top of a tower from a point was observed to be 45°.
On walking 30m away from that point, it was found to be 30°. Find the height of
the house.

5. (a) From a helicopter flying vertically above a straight road, the angles of depressions
of two consecutive kilometer stone on the same side are found to be 45° and 60°.
Find the height of the helocopter.

(b) From the top of a tower of 100m, the angle of depression of two places due to east
of it are respectively 45° and 60°. Find the distance between the two places.

(c) From the top of a tower of 200 m, the angle of depression of two boats which are
in a straight line on the same side of the tower are found to be 30° and 45°. Find
the distance between the boats.

6. (a) From the top of 21 metre high cliff, the angles of depression of the top and bottom
of a towers are observed to be 45° and 60° respectively. Find the height of the
tower.

(b) From the top of a cliff the measure of the angle of depression of the top and bottom
of a building are found to be 30° and 45°. If the height of the cliff is 100 metres.
Find the height of the building.

7. (a) Two poles stand on either side of a road. At the point midway between the two
posts, the angles of inclination(elevation) of their tops are 30° and 60°. Find the
length of the shorter posts if the longer post is of 15m long.

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(b) Two lamp posts are of equal height. A girl standing mid way between them observes
the angle of elevation of either posts to be 45°. After walking 15m towards one of
them, she observes its elevation to be 60°. Find the heights of the posts and the
distance between them.

(c) The posts are 120m apart, and the height of one is double that of other. From the
middle point of the line joining the feet, an observer found the angle of elevation
of their tops to be complementary. Find the height of the longer post.

8. (a) A flagstaff stands upon the top of a building. The angles of elevation of the top
of the building and flagstaff are observed to be 45° and 60° respectively. If the
distance between the observer and the building is 10m, find the length of the
flagstaff.

(b) A flagstaff stands on the top of a tower. The angles subtended by the tower and the
flagstaff at a point 100 metres away from the foot of the tower are found to be 45°
and 15°. Find the length of the flagstaff.

(c) A flagstaff is placed at one corner of a rectangular garden 40m long and 30m wide.
If the angle of elevation of the top of the flagstaff from the opposite corner is 30°.
Find the height of the flagstaff.

9. (a) From the top and bottom of a tower, the angle of depression of the top of the house
and angle of elevation of the house are found to be 60° and 30° respectively. If the
height of the building is 20m. find the height of the tower.

(b) From the roof of a house 30m height, the angle of elevation of the top of a tower is
45° and the angle of depression of its foot is 30°. Find the height of the tower.

(c) A crow is sitting on the top of a tree which is in front of a house. The angle of
elevation and angle of depression of the crow from the bottom and top of the
house are 60° and 30° respectively. If the height of the tree is 21m. Find the height
if the house.

10. (a) A ladder of 18m reaches a point of 18m below the top of vertical flagstaff. From the
foot of the ladder the angle elevation of the flagstaff is 60°. Find the height of the
flagstaff.

(b) A ladder 15m long reaches a point 15 below the top of a vertical flagstaff. From
the foot of the ladder, the angle of elevation of the top of flagstaff is 60°. Find the
height of the flagstaff.

11. (a) A pole is divided by a point in the ratio 1:9 from bottom to top. If the two parts
of the pole subtend equal angles at 20m away from the foot of the pole, find the
height of the pole.

(b) AB is a vertical pole with its foot B on a level of ground. A point C on AB divides
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.

12. (a) A man 1.75m stands at a distance of 8.5m from a lamp post and it is observed that
his shadow is 3.5m long. Find the height of the lamp post.

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(b) A tree of 12m height is broken due to wind such a way that its top touches the
ground and makes an angle of 60° with the ground. At what height from the bottom
is the tree broken by the wind.

13. The angle of elevation of an aeroplane from a point in the ground is 45°. After 15 seconds
of flight the angle if elevation changes to 30°. If the aeroplane is flying horizontally at a
height of 4000m in the same direction, find the speed of aeroplane.

14. The angle of elevation of the top of a tower is 45° from a point 10m above the water
level of a lake. The angle of depression of its image in the lake from the same point is
60°. Find the height of the tower above the water level.

15. A rope dancer was walking on a loose rope tied to the tops of two posts each 8m high.
When the dancer was 2.4m above the ground, it was found that the stretched pieces of
the rope made angles of 30° and 60° with the horizontal line parallel to the ground. Find
the length of the rope.

Project Work

16. Select buildings around your school. Find the angle of elevation by using sextants
or theodolite. Find the height of the buildings taking the distance between the two
places in meter.

17. Write a short note on " Height and Distance" with its application.

2. (a) 30 3 m (b) 20 3 m

3. (a) x = y = 27.32 m (b) x = 200 m, y = 115.47 m

(c) x = 7.79 m, y = 4.5 m (d) x = 20 3, y = 40 m

4. (a) 51.96 m (b) 40.98 m 5.(a) 2365.98 m

(b) 42.26 m (c) 146.4 m 6.(a) 8.87 m (b) 42.27 m

7. (a) 5 m (b) 35.49 m, 70.98 m (c) 84.85 m

8. (a) 7.32 m (b) 73.2 m (c) 28.86 m

9. (a) 80 m (b) 81.96 m (c) 28 m

10. (a) 27 m (b) 22.5 m 11.(a) 178.88 m (b) 22.36 m

12. (a) 6 m (b) 5.57 m

13. 195.2 m/s 14. 37.32 m 15. 17.67 m

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10Vectors

10.0 Review

Discuss the following questions:
a. Classify the following quantities to vectors or scalars with your reasons.

(i) Temperature (ii) Pressure (iii) Force (iv) Length

(v) Area (vi) Acceleration (vii) Displacement (viii) Time

b. Give an example of each of the following vectors:

(i) Column vector (ii) Row vector (iii) unit vector

(iv) Equal vectors (v) Negative vectors (vi) position vectors

(vii) Unequal vectors (viii) Like and unlike vectors.

c. What role do vectors play in our daily life ? Give some examples.

o o oo oo
d. Let a = (2, -3) and b = (1, 2), find a + b and a - b . Express them in the
oo
form of x i + y j .

10.1 Scalar (or Dot) Product of Two Vectors

oo
Let OA = (2, 3), OB = (-3, 2) be two vectors. Plot them in a graph paper. Then,

(a) Find the product of x-components and y-components Y
of the vectors.

(b) Add the products.

(c) What is the result ? BA X
oo X' O

(d) Are OA and OB perpendicular to each other?
Here, 2 × (-3) + 3 × 2 = - 6 + 6 = 0

oo
OA is perpendicular to OB .

There are two types of product of two vectors: Y'
(a) Scalar product or dot product.
(b) Vector product or cross product.

We discuss only about scalar product or dot product of two vectors.

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Scalar Product of Two Vectors

The scalar product of two vectors oa and ob is defined as the
product of the magnitude of two vectors multiplied by the cosine
of the angle T between their directions.

Thus, oa · ob = |oa | |ob | cosT = ab cosT
where, |oa |= a and |ob |= b.

AO = oa and OB = ob

Now, draw perpendicular BM from B to OA.

Now, oa · ob = |oa | |ob |cosT = a b cos T

= (OA) (OB) cos T

= OA (OB cos T)
= (OA) (OM) = (magnitude of oa ) (component of ob in the direction of oa )

So, it is clear that the scalar product of two vectors is equivalent to the product of the
magnitude of one vector with the component of the other vector in the direction of this
vector.

If we write oa · ob , the rotation of oa towards ob is anti-clockwise and the angle T is taken to
be positive.

? oa · ob = ab cosT

If we write ob ;oa , the rotation of ob towards oa is clockwise and the angle T is taken to be
negative.

? ob · oa = b a cos(–T) = ba cosT
Hence, oa · ob = ob · oa

Note :
Let oa = x1oi + y1oj and ob = x2oi + y2oj be two vectors in terms of the components. Then
we know that oi and oj are unit vectors along X-axis and Y-axis respectively.

oi . oj = |oi ||oj |cos 90° = 0, oi . oi = |oi | |oi | cos0° = 1

Now, oa . ob = (x1 oi + y1) . (x2oi + y2oj ) = x1oi .(x2oi + y2oj ) + y1 oj . (x2oi + y2 oj )
= x1x2 oi . oi + x1y2 oi . oj + y1 x2 oj . oi + y1y2 oj . oj = x1x2 + y1y2

? oa . ob = x1x2 + y1y2

Hence, the scalar product of two vectors is equal to the sum of the product of their
corresponding components.

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Two Important Results of Scalar Product of Two Vectors.

(a) If oa and ob are perpendicular to each other, i.e., T = 90° B A
Then, oa . ob = |oa | |ob | cosT
= |oa | |ob | cos90° = 0 o
Hence, oa . ob = 0 b

o
a

Conversely, of oa . ob = 0, then T = 90°

oo
(b) If a and b are parallel to each other,

i.e., T = 0° or T = 180°

o o |oa | |ob | cos0° = 1
a. b=

and o o = |oa | |ob | cos180° = ab.(-1) = - ab
a. b

oo oo
Conversely, if a . b = ± ab, T = 0° or 180°, a and b are parallel to each other.

Some Important Results

a. If |oa | = a and |ob | = b, oo
a . b = abcosT.

oo (i) |oa |= 0 (ii) |ob | = 0 or o o
b. If a . b = 0, a A b

oo
c. If T = 0°, the value of a . b = ab is maximum.

oo
d. If T = 180°, the value of a . b = - ab is minimum.

oo oo
e. a . b = b . a .

oo o o oo
f. a . (k b ) = (k a ) . b = k( a . b ), where k z 0. k  R.

g. o o = |oa | |oa | cos0 = a.a. =a2
a. a

Also,

(i) o o = |oi | |oi | cos0° = 1.1.1 = 1
i. i

(ii) o o = |oj | |oj | cos0° = 1.1.1 = 1
j. j

(iii) o o = |oi | |oj | cos90° = 1.1.0 = 0
i. j

o . o = |oj | |oi | cos90° = 1.1.0 = 0
(iv) j i

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Angle between Two Vectors

Let us consider two points A(a1, a2) and B (b1, b2) in the plane. Then,

position vector of A = OA = oa = a1
a2
= ob = b1
Position vector of B = OB b2

Magnitudes of oa and ob are

| OA | = OA = a = |oa |

| OB | = OB = b = |ob |

Let ‘XOA = E, ‘XOB = D and ‘AOB = T. Then, ‘AOB = D – E = T.

Draw perpendiculars AM and BN from A and B to the x-axis. Then,

OM = a1, MA = a2, ON = b1 and NB = b2.

From the right-angled triangle OMA,

cosE = OM = aa1 ? a1 = a cos E
OA ? a2 = a sinE
aa2
sinE = MA =
OA

Similarly from the right-angled triangle ONB,

b1 = b cos D and b2 = b sin D
Now, a1b1 + a2b2 = a cos E b cosD + a sin E b sin D

= ab cos(D – E) = |oa | |ob | cos T ................(i)
But, by the definition of scalar product of two vectors, oa and ob

oa · ob = | oa | |ob | cos T .............(ii)

Now, from (i) and (ii) oa · ob = a1b1 + a2b2.

This result leads us to define scalar product of two vectors in another way.

Let oa = a1 and oa = b1 be two vectors. Then the scalar product of oa and ob is denoted
a2 b2
by oa · ob and is defined by oa · ob = a1 b1
a2 · b2 = a1b1 + a2b2.

Again from (i), |oa | |ob | cos T = a1b1 +a2b2

or, cos T = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |
oa ob .
This result gives us angle between two vectors and

cosT = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |

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Worked out Examples

oo oo
Example 1. (a) If a = (4,5) and b = (2, 3), find the scalar product of a and b .
Solution:
(b) If |oa |= 4 and o = 3, T = 45°. find o o .
Example 2. b a. b
Solution:
oo
Example 3. (a) Here, a = (4, 5), b = (2, 3)
Solution:
Since (a1, a2). (b1,b2) = a1b1 + a2b2
oo
a . b = (4, 5).(2,3) = 4.2 + 5.3 = 8+15= 23

oo
(b) Here, | a | = 4, | b | = 3, T = 45°

oo o o 1 = 6 2
a . b = | a | | b | cosT = 4.3 cos45° = 12. 2

o oo ooo
If a = 2 i + j and b = i - 2 j , then, find the following:

oo oo
(i) a . b (ii) angle between a and b

o o oo o o
Here, a = 2 i + j , b = i - 2 j

oo o o o o
(i) a . b = (2 i + j ). ( i - 2 j ) = 2.1 + 1.(-2) = 2 - 2 = 0

oo oo
Since a . b = 0, a and b are perpendicular to each other.

o
(ii) Here,| a | = 22 + 12 = 4 + 1 = 5

o
| b | = 12+(-2)2 = 1 + 4 = 5

oo
Let T be the angle between a and b , we get,

cosT = oo = 0 =0
a. b 55
| oa || ob |

? T = 90°

ooo o oo o
(a) If a = 3 i + 4 j and b = -8 i + 6 j , show that a is perpendicular
oo o oo
to b , (b) If a = (2, 4) and b = (4, 8), show that a and b are parallel to

each other.

o o oo oo
(a) Here, a = 3 i + 4 j , b = - 8 i + 6 j

oo oo oo
a . b = (3 i + 4 j ) . (-8 i + 6 j )

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= 3.(-8) + 4.6 = - 24 + 24 = 0

oo o o
Since a . b = 0, a and b are perpendicular to each other.

( ) ( )(b)
o 2 o 4
Here, a = 4 , b= 8

( ) ( )o o 2 . 4 = 2.4 + 4.8 = 8 + 32 = 40
4 8
a. b =
o
| a | = 22 + 42 = 4 + 16 = 20 = 2 5

o
| b | = 42 +82 = 16+64 = 80 = 4 5

oo
Let T be the angle between a and b

Now, cosT oo =2 40 5 = 1 = cos0°
= a. b 5 ×4

oo
| a || b |

? T = 0°

oo
Hence, a and b are parallel to each other.

oo oo
Example 4. (a) Find the value of m if two vectors m i + 7 j and 4 i + 8 j are
Solution:
perpendicular to each other.

( ) ( )o

(b) If OA =
2 o -3 and ‘AOB = 90°, find the value of m.
m and OB = 4

(a) Here, let o = o + o o oo
a mi 7 j, b =4 i+8 j

oo oo
Since a and b are at right angle, a . b = 0

oo oo oo
i.e. a . b = (m i + 7 j ) . (4 i + 8 j )

or, 0 = 4m + 56

or, 4m = - 56 ? m = -14

( ) ( )(b) o o
oo 2 , OB = b= -3 , ‘AOB = T = 90°
Let OA = a = m 4

Then, by definition of dot product,

( ) ( )o o 2 . -3 = 2(-3) + m.4 = - 6 + 4m
m 4
a. b= oo

Since, T = 90° or, a . b = 0

or, - 6 + 4m = 0

or, 4m = 6 ? m= 3
2

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