Trigonometry – Exact PDF Typing
Same-to-same reconstruction of the supplied 66-page handwritten PDF. Page order, question numbering, working steps, formulas, source answers and unusual source calculations are preserved.
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PDF Pages 1–12
Source page order preserved exactly.
→ In ΔBCD
tan 45° = n / BD
or, 1 = n / 5
∴ n = 5.
In ΔABD.
tan θ = (AC + n) / BD
or, tan θ = (3.66 + 5) / 5
or, tan θ = 8.66 / 5
or, tan θ = tan−1(1.732).
∴ θ = 60°.
9)
AB is a vertical pole with its foot B on a leveled 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 20 m from the foot of the pole, find the height of the pole.
tan θ = 2n / 20
tan 2θ = (3n + 2n) / 20
or, 2tan θ / (1 − tan²θ) = 5n / 20
or, 2 × (2n/20) / [1 − (2n/20)²] = 5n / 20
or, (n/5) / [1 − 4n²/400] = 5n / 20
or, (n/5) / [(400 − 4n²)/400] = 5n / 20
or, 80n / (400 − 4n²) = 5n / 20
or, 80 / (400 − 4n²) = 1 / 4
or, 400 − 4n² = 320
or, 4n² = 400 − 320
or, 4n² = 80
or, n² = 80 / 4
or, n² = 20
or, n = √20
∴ n = 4.47 m.
3n + 2n = 5n
= 5 × 4.47
= 22.36.
10)
A ladder 20 m long leans against a house situated at one side of a road at an angle of 30° with the ground. When it is turned so that it rests against another house on the other side of the road it makes an angle of 45° with the house. Find the width of the road.
→ In ΔABC.
cos 30° = b / h
or, √3 / 2 = BC / 20
or, 2BC = 20√3
or, BC = 20√3 / 2
or, BC = 10√3 m.
In ΔCDE.
cos 45° = b / h
or, 1/√2 = EC / 20
or, √2 EC = 20
or, EC = 20 / √2
or, EC = 10√2.
Width of the road = BC + EC
= 10√3 + 10√2
= 17.3205 + 14.1421
= 31.4626 m.
11)
Two poles stand on either side of a road. At the point mid-way b/tw the two posts the angles of elevation of their tops are 30° and 60°. Find the length of the shorter post if the other post is 15 m longer.
→ In ΔECD.
tan 60° = p / b
or, tan 60° = 15 / CD
or, √3 = 15 / CD
or, CD = 15 / √3
∴ CD = 5√3.
In ΔABC.
tan 30° = p / b
or, 1/√3 = AB / BC
or, 1/√3 = AB / 5√3
or, AB = 5√3 / √3
∴ AB = 5.
Therefore, the length of the shorter pole is 5 m.
12)
The angle of elevation of an aeroplane flying horizontally at a height of 750 m above the ground is observed to be 60°. After 5 seconds, the angle of elevation is observed to be 30°. Find the speed of the aeroplane in km per hour.
In ΔABC.
tan 60° = 750 / BC
or, √3 = 750 / BC
or, BC = 750 / √3
∴ BC = 433.01 m.
In ΔBED.
tan 30° = 750 / BE
or, 1/√3 = 750 / BE
or, 1/√3 = 750 / (433.01 + CE)
or, 433.01 + CE = 750√3
or, 433.01 + CE = 1299.03
or, CE = 1299.03 − 433.01
or, CE = 866.02 m.
distance = 866.03 m
= 0.86603 km.
Time = 5 second
= 5 / (60 × 60)
= 0.001388 hr.
speed = distance travelled / Time
= 0.86603 / 0.001388
= 623.3909 m/s. [as written in source]
16)
If the length of the shadow of a vertical column increases by 10√3 m when the altitude of the sun becomes 45° from 60°. Find the height of the column and the length of the shadow when the sun’s altitude was 60°.
In ΔABC.
tan 60° = AB / BC
or, √3 = AB / BC
or, AB = √3 BC …(i)
In ΔABD.
tan 45° = AB / BD
or, 1 = √3 BC / (BC + 10√3)
or, √3 BC = BC + 10√3
or, 1.7320 BC − BC = 10√3
or, 0.7320 BC = 10√3
or, BC = 10√3 / 0.7320
∴ BC = 23.66 m.
Now, AB = √3 BC
= √3 × 23.66
= 40.98025 m.
13)
A rope dancer was walking on a loose rope tied to the tops of two posts, each 8 m high. When the dancer was 2.4 m above the ground, it was found that the shorter pieces of the rope made angles of 30° and 60° with the horizontal line parallel to the ground. Find the length of the rope.
→ In ΔDEF.
sin 30° = 5.6 / DE
or, 1/2 = 5.6 / DE
∴ DE = 11.2 m.
In ΔBEF.
sin 60° = 5.6 / BE
or, 1/√2 = 5.6 / BE [as written in source]
or, BE = 7.9195
length of rope = 7.9195 + 11.2
= 19.1195 m.
14)
The angle of elevation of the top of a tower is 45° from a point 10 m above water level of a lake. The angle of depression to its image in the lake is 60°. Find the height of the tower above water level.
→ In ΔBFD.
tan 45° = n / FD
or, 1 = n / FD
∴ FD = n …(i)
In ΔFED.
tan 60° = (n + 20) / FD
or, √3 = (n + 20) / n
or, √3 n = n + 20
or, 0.73n = 20
or, n = 20 / 0.73
n = 27.39 m.
height of tower = 27.39 + 10
= 37.39 m.
Exercise 9.5
Starts on physical PDF page 13.
Exercise 9.5
B) Short type questions
1(a) sin θ = cos θ
sin θ = sin(90° − θ)
θ = 90° − θ
or, 2θ = 90°
∴ θ = 45°.
(b) tan α = cot α
tan α = tan(90° − α)
or, α = 90° − α
or, 2α = 90°
or, α = 45°
tan α = tan(270° − α)
or, α = 270° − α
or, 2α = 270°
or, α = 135°
(c) cos 2n = sin n
cos 2n = cos(270° + n)
or, 2n = 270° + n
or, 2n − n = 270°
∴ n = 270°.
cos 2n = cos(90° + n)
or, 2n = 90° + n
or, n = 90°.
(d) sin 3θ = −cos θ
sin 3θ = sin(270° − θ)
or, 3θ = 270° − θ
or, 4θ = 270°
∴ θ = 67.5°
Again,
sin 3θ = sin(270° + θ)
or, 3θ = 270° + θ
or, 2θ = 270°
∴ θ = 135°.
(e) tan n = cot 5n
tan n = tan(270° − 5n)
or, n = 270° − 5n
or, 6n = 270°
∴ n = 45°
Again,
tan n = tan(90° − 5n)
or, n = 90° − 5n
or, 6n = 90°
∴ n = 15°.
(f) cos 3n = sin 2n
cos 3n = cos(90° − 2n)
or, 3n = 90° − 2n
or, 5n = 90°
∴ n = 18°.
Again,
cos 3n = cos(270° + 2n)
or, 3n = 270° + 2n
or, n = 270°.
(g) sin 3n = cos 7n
cos(270° + 3n) = cos 7n
or, 270° + 3n = 7n
or, 270° = 4n
∴ n = 67.5°
Again,
sin 3n = sin(90° − 7n)
or, 3n = 90° − 7n
or, 10n = 90°
∴ n = 9°.
3(a) 2cos²θ = −√3 cos θ
2cos²θ + √3 cosθ = 0
cosθ(2cosθ + √3) = 0
Either,
cosθ = 0
cosθ = cos90°
θ = 90°.
OR,
2cosθ + √3 = 0
2cosθ = −√3
cosθ = −√3/2
cosθ = cos(180° − 30°)
θ = 150°
cosθ = cos(180° + 30°)
θ = 210°.
(b) 2cos²θ − √3 cosθ = 0
cosθ(2cosθ − √3) = 0
Either,
cosθ = 0
θ = 90°
OR,
2cosθ − √3 = 0
cosθ = √3/2
cosθ = cos30°
∴ θ = 30°.
(c) 2cos²θ = 3sin θ
2cos²θ − 3sinθ = 0
2(1 − sin²θ) − 3sinθ = 0
2 − 2sin²θ − 3sinθ = 0
2sin²θ + 3sinθ − 2 = 0
2sinθ(sinθ + 2) − 1(sinθ + 2) = 0
(2sinθ − 1)(sinθ + 2) = 0
Either,
2sinθ − 1 = 0
sinθ = 1/2
sinθ = sin30°
∴ θ = 30°.
OR,
sinθ = −2
rejected.
(d) cos²(θ/2) − cos(θ/2) + 1/4 = 0
4cos²(θ/2) − 4cos(θ/2) + 1 = 0
2cos(θ/2)[2cos(θ/2) − 1] − 1[2cos(θ/2) − 1] = 0
[2cos(θ/2) − 1][2cos(θ/2) − 1] = 0
or, 2cos(θ/2) − 1 = 0.
or, 2cos(θ/2) − 1 = 0
or, cos(θ/2) = 1/2
or, cos(θ/2) = cos60°
∴ θ/2 = 60°
cos(θ/2) = cos(180° − 60°)
cos(θ/2) = cos120°
∴ θ/2 = 120°.
C) Long type question
1) Solve the following equations (0° ≤ θ ≤ 360°)
(a) 4cos²θ + 4sinθ = 5
4(1 − sin²θ) + 4sinθ − 5 = 0
or, 4 − 4sin²θ + 4sinθ − 5 = 0
or, −4sin²θ + 4sinθ − 1 = 0
or, (4sin²θ − 4sinθ + 1) = 0
or, (2sinθ − 1)² = 0
or, 2sinθ − 1 = 0.
or, sinθ = 1/2
or, sinθ = sin30°
∴ θ = 30°
Again,
sinθ = sin(180° − 30°)
θ = 150°.
(b) cos²θ + 3sin²θ + sinθ = 2
1 − sin²θ + 3sin²θ + sinθ = 2
or, 1 + 2sin²θ + sinθ = 2
or, 2sin²θ + sinθ − 1 = 0
or, 2sin²θ + 2sinθ − sinθ − 1 = 0
or, 2sinθ(sinθ + 1) − 1(sinθ + 1) = 0
or, (2sinθ − 1)(sinθ + 1) = 0
Either,
2sinθ − 1 = 0
sinθ = 1/2
sinθ = sin30°
∴ θ = 30°.
OR,
sinθ + 1 = 0
sinθ = −1
sinθ = sin270°
∴ θ = 270°.
(c) sin²θ − 2cosθ = −1/4
1 − cos²θ − 2cosθ = −1/4
or, cos²θ + 2cosθ − 1¼ = 0
or, 4cos²θ + 8cosθ − 5 = 0
or, 4cos²θ + 10cosθ − 2cosθ − 5 = 0
or, 2cosθ(2cosθ + 5) − 1(2cosθ + 5) = 0
or, (2cosθ − 1)(2cosθ + 5) = 0
Either,
2cosθ − 1 = 0
cosθ = 1/2
cosθ = cos60°
∴ θ = 60°
Again, cosθ = cos(360° − 60°)
∴ θ = 300°.
OR,
2cosθ + 5 = 0
cosθ = −5/2
Rejected.
(d) cos²θ − sinθ = 1/4
1 − sin²θ − sinθ = 1/4
or, sin²θ + sinθ + 1/4 − 1 = 0
or, sin²θ + sinθ − 3/4 = 0
or, 4sin²θ + 4sinθ − 3 = 0
or, 4sin²θ + 6sinθ − 2sinθ − 3 = 0
or, 2sinθ(2sinθ + 3) − 1(2sinθ + 3) = 0
or, (2sinθ − 1)(2sinθ + 3) = 0
Either,
2sinθ − 1 = 0
sinθ = 1/2
sinθ = sin30°
∴ θ = 30°.
OR,
2sinθ + 3 = 0
sinθ = −3/2
rejected.
(e) 2√3 sin²θ = cosθ
2√3(1 − cos²θ) − cosθ = 0
2√3 − 2√3 cos²θ − cosθ = 0
or, 2√3 cos²θ + cosθ − 2√3 = 0
or, 2√3 cos²θ + 4cosθ − 3cosθ − 2√3 = 0
or, 2cosθ(√3 cosθ + 2) − √3(√3 cosθ + 2) = 0
or, (√3 cosθ + 2)(2cosθ − √3) = 0
Either,
√3 cosθ + 2 = 0
cosθ = −2/√3
OR,
2cosθ − √3 = 0
cosθ = √3/2
cosθ = cos30°
∴ θ = 30°
Again, cosθ = cos(360° − 30°)
θ = 330°.
(f) 2cos²θ − cosθ − 1 = 0
2cos²θ − 2cosθ + cosθ − 1 = 0
or, 2cosθ(cosθ − 1) + 1(cosθ − 1) = 0
or, (2cosθ + 1)(cosθ − 1) = 0.
Either,
2cosθ + 1 = 0
cosθ = −1/2
cosθ = cos120°
θ = 120°.
OR,
cosθ − 1 = 0
cosθ = 1
cosθ = cos0°
∴ θ = 0°
cosθ = cos(360° + 0°)
θ = 360°.
(g) tan²θ − 3secθ + 3 = 0
sin²θ / cos²θ − 3/cosθ + 3 = 0
or, sin²θ − 3cosθ + 3cos²θ = 0
or, 1 − cos²θ − cosθ + 3cos²θ = 0 [as written in source]
or, 2cos²θ − cosθ + 1 = 0 [as written in source]
or, 2cos²θ − 2cosθ − cosθ + 1 = 0
or, 2cosθ(cosθ − 1) − 1(cosθ − 1) = 0
or, (2cosθ − 1)(cosθ − 1) = 0.
Either,
2cosθ − 1 = 0
cosθ = 1/2
cosθ = cos60°
θ = 60°
cosθ = cos(360° − 60°)
θ = 300°.
OR,
cosθ − 1 = 0
cosθ = 1
cosθ = cos0°
θ = 0°
cosθ = cos(360° + 0°)
θ = 360°.
(h) cot²θ + 3/sinθ + 3 = 0
cos²θ / sin²θ + 3/sinθ + 3 = 0
or, cos²θ + 3sinθ + 3sin²θ = 0
or, 1 − sin²θ + 3sinθ + 3sin²θ = 0
or, 2sin²θ + 3sinθ + 1 = 0
or, 2sin²θ + 2sinθ + sinθ + 1 = 0
or, 2sinθ(sinθ + 1) + 1(sinθ + 1) = 0
or, (sinθ + 1)(2sinθ + 1) = 0
Either,
sinθ + 1 = 0
sinθ = −1
sinθ = sin(180° + 90°)
θ = 270°
Again, sinθ = sin(360° − 90°)
θ = 270°.
OR,
2sinθ + 1 = 0
sinθ = −1/2
sinθ = sin(180° + 30°)
θ = 210°
Again, sinθ = sin(360° − 30°)
θ = 330°.
(i) 1 + cos2θ = cosθ
1 + (2cos²θ − 1) = cosθ
or, 2cos²θ = cosθ
or, 2cos²θ − cosθ = 0.
or, cosθ(2cosθ − 1) = 0
Either,
cosθ = 0
cosθ = cos0°
∴ θ = 0°. [as written in source]
OR,
2cosθ − 1 = 0
cosθ = 1/2
cosθ = cos60°
θ = 60°
Again, cosθ = cos(360° − θ)
θ = 300°.
(j) cot²θ + (√3 + 1/√3)cotθ = −1
cot²θ + √3cotθ + (1/√3)cotθ + 1 = 0
or, √3cot²θ + 3cotθ + cotθ + √3 = 0
or, √3cotθ(cotθ + √3) + 1(cotθ + √3) = 0
or, (√3cotθ + 1)(cotθ + √3) = 0.
Either,
cotθ = −√3
cotθ = cot(180° − 30°)
θ = 150°
Again, cotθ = cot(360° − 30°)
θ = 330°.
OR,
cotθ = −1/√3
cotθ = cot(180° − 60°)
θ = 120°
Again, cotθ = cot(360° − 60°)
θ = 300°.
(k) 4cos²θ + √3 = 2(√3 + 1)cosθ
4cos²θ + √3 = 2√3cosθ + 2cosθ
or, 4cos²θ − 2√3cosθ − 2cosθ + √3 = 0
or, 2cosθ(2cosθ − √3) − 1(2cosθ − √3) = 0
or, (2cosθ − √3)(2cosθ − 1) = 0
Either,
2cosθ − √3 = 0
cosθ = √3/2
θ = 30°
Again, θ = 330°.
OR,
2cosθ − 1 = 0
cosθ = 1/2
θ = 60°
Again, θ = 300°.
(l) sin2θ + cosθ = 1 + 2sinθ
2sinθcosθ + cosθ − 1 − 2sinθ = 0
or, cosθ(2sinθ + 1) − 1(2sinθ + 1) = 0
or, (2sinθ + 1)(cosθ − 1) = 0
Either,
2sinθ + 1 = 0
sinθ = −1/2
sinθ = sin(180° + 30°)
θ = 210°
Again, sinθ = sin(360° − 30°)
θ = 330°.
OR,
cosθ − 1 = 0
cosθ = 1
cosθ = cos0°
∴ θ = 0°.
(m) cotθ + tanθ = 2secθ
cosθ/sinθ + sinθ/cosθ = 2/cosθ
or, (cos²θ + sin²θ)/(sinθ·cosθ) = 2/cosθ
or, 1/sinθ = 2.
or, 2sinθ = 1
or, sinθ = 1/2
or, sinθ = sin30°
∴ θ = 30°
Again,
sinθ = sin(180° − 30°)
θ = 150°.
2) Solve the following equations (0° ≤ n ≤ 360°)
(a) sin n + cos n = √2
sin n = √2 − cos n
Squaring both sides:
sin²n = (√2 − cosn)²
or, sin²n = 2 − 2√2 cosn + cos²n
or, 1 − cos²n − 2 + 2√2 cosn − cos²n = 0
or, −2cos²n + 2√2 cosn − 1 = 0
or, 2cos²n − 2√2 cosn + 1 = 0.
or, 2cos²n − √2cosn − √2cosn + 1 = 0
or, √2cosn(√2cosn − 1) − 1(√2cosn − 1) = 0
or, (√2cosn − 1)(√2cosn − 1) = 0
or, (√2cosn − 1)² = 0
or, √2cosn − 1 = 0
or, cosn = 1/√2
or, cosn = cos45°
∴ n = 45°
Again, cosn = cos(360° − 45°)
n = 315°.
(b) √3 sin n + cos n = 1
√3 sin n = 1 − cos n
Squaring both sides:
3sin²n = 1 − 2cosn + cos²n
or, 3 − 3cos²n = 1 − 2cosn + cos²n
or, 1 − 2cosn + cos²n + 3cos²n − 3 = 0.
or, 4cos²n − 2cosn − 2 = 0
or, 4cos²n − 4cosn + 2cosn − 2 = 0
or, 4cosn(cosn − 1) + 2(cosn − 1) = 0
(cosn − 1)(4cosn + 2) = 0
Either,
cosn − 1 = 0
cosn = 1
cosn = cos0°
∴ n = 0°.
OR,
4cosn + 2 = 0
cosn = −2/4 = −1/2
cosn = cos(180° − 60°)
n = 120°
Again, cosn = cos(180° + 60°)
n = 240°.
(c) cos n − √3 sin n = 1
√3 sin n = cos n − 1
Squaring both sides:
3sin²n = cos²n − 2cosn + 1.
or, 3 − 3cos²n = cos²n − 2cosn + 1
or, 4cos²n − 2cosn − 2 = 0
or, 4cos²n − 4cosn + 2cosn − 2 = 0
or, 4cosn(cosn − 1) + 2(cosn − 1) = 0
or, (cosn − 1)(4cosn + 2) = 0
Either,
cosn − 1 = 0
cosn = 1
cosn = cos0°
∴ n = 0°
Again, cosn = cos(360° − 0°)
∴ n = 360°.
OR,
cosn = −2/4
cosn = −1/2
cosn = cos(180° − 60°)
n = 120°
Again, cosn = cos(180° + 60°)
n = 240°.
(e) √3 cos n + sin n = √3
√3 cos n = √3 − sin n
Squaring both sides:
3cos²n = 3 − 2√3 sinn + sin²n
or, 3 − 3sin²n = 3 − 2√3 sinn + sin²n
or, 3 − 2√3 sinn + sin²n + 3sin²n − 3 = 0
or, 4sin²n − 2√3 sinn = 0
or, 2sinn(2sinn − √3) = 0
Either,
2sinn = 0
sinn = 0
sinn = sin0°
∴ n = 0°.
OR,
sinn = √3/2
sinn = sin60°
∴ n = 60°
Again, sinn = sin(90° + 60°)
∴ n = 150°. [as written in source]
(c) √3 sin n − cos n = √2
Here, √3 sinn − cosn = √2
Dividing both sides by √[(coeff. of cosn)² + (coeff. of sinn)²]
√[(-1)² + (√3)²] = √(1 + 3) = 2
Now, (√3/2)sinn − (1/2)cosn = √2/2
or, sin60°·sinn − cos60°·cosn = 1/√2
or, −(cos60°·cosn − sin60°·sinn) = 1/√2
or, −cos(n + 60°) = 1/√2
or, cos(n + 60°) = −1/√2
or, cos(n + 60°) = cos(180° + 45°)
or, cos(n + 60°) = cos225°
n = 225° − 60°
n = 165°.
cos(n + 60°) = −1/√2
cos(n + 60°) = cos(180° − 45°)
n + 60° = 135°
∴ n = 75°.
(g) cos n + √3 sin n = 2
√3 sinn = 2 − cosn
Squaring both sides:
(√3 sinn)² = (2 − cosn)²
or, 3sin²n = 4 − 4cosn + cos²n
or, 3 − 3cos²n = 4 − 4cosn + cos²n
or, 4 − 4cosn + cos²n + 3cos²n − 3 = 0
or, 4cos²n − 4cosn + 1 = 0
or, (2cosn − 1)² = 0
or, 2cosn − 1 = 0
or, cosn = 1/2
or, cosn = cos60°
∴ n = 60°
Again, cosn = cos(360° − 60°)
∴ n = 300°.
(h) cos n + (1/√3) sin n = 1
sinn/√3 = 1 − cosn
Squaring both sides:
sin²n/3 = (1 − cosn)²
or, 1 − cos²n = 3 − 6cosn + 3cos²n
or, 3 − 6cosn + 3cos²n − 1 + cos²n = 0
or, 4cos²n − 6cosn + 2 = 0
or, 2cos²n − 3cosn + 1 = 0
or, 2cos²n − 2cosn − cosn + 1 = 0
or, 2cosn(cosn − 1) − 1(cosn − 1) = 0
or, (2cosn − 1)(cosn − 1) = 0
Either,
2cosn − 1 = 0
cosn = 1/2
cosn = cos60°
n = 60°
Again, cosn = cos(360° − 60°)
n = 300°.
OR,
cosn − 1 = 0
OR,
cosn − 1 = 0
or, cosn = 1
or, cosn = cos0°
∴ n = 0°.
(i) sin n + cos n = 1
sinn = 1 − cosn
Squaring both sides:
sin²n = 1 − 2cosn + cos²n
or, 1 − cos²n = 1 − 2cosn + cos²n
or, 1 − 1 − 2cosn + 2cos²n = 0
or, 2cosn(cosn − 1) = 0
Either,
2cosn = 0
cosn = 0
cosn = cos90°
∴ n = 90°.
OR,
cosn − 1 = 0
cosn = 1
cosn = cos0°
∴ n = 0°.
3) Solve the following equations
(a) sin3n + sin n = 2sin n (0° ≤ n ≤ 180°)
sin3n + sinn − 2sinn = 0
or, sin3n − sinn = 0
or, 2cos[(3n+n)/2] · sin[(3n−n)/2] = 0
or, 2cos2n · sinn = 0
or, 2(1 − 2sin²n) · sinn = 0
Either,
2 − 4sin²n = 0
sin²n = 2/4
sinn = √(2/4) = 1/√2
sinn = sin45°
∴ n = 45°
sinn = sin(180° − 45°)
n = 135°.
OR,
sinn = 0
sinn = sin0°
n = 0°.
(b) cos3n + cos n = 2cos n (0° ≤ n ≤ 360°)
cos3n + cosn − 2cosn = 0
or, cos3n + cosn(1 − 2) = 0
or, cos3n + cosn = 0 [as written in source]
or, 2cos[(3n+n)/2] · cos[(3n−n)/2] = 0
or, 2cos2n · cosn = 0
or, 2(1 − cos²n) · cosn = 0 [as written in source]
or, (2 − 2cos²n)cosn = 0
Either,
2 − 2cos²n = 0
cos²n = 1
cosn = √1
cosn = 1
cosn = cos0°
∴ n = 0°
cosn = cos(180° + 180°)
cosn = cos360°
∴ n = 360°.
OR,
cosn = 0
cosn = cos90°
n = 90°
cosn = cos(180° + 90°)
cosn = cos270°
∴ n = 270°.
(c) cos3θ + cosθ = cos2θ
2cos[(3θ+θ)/2] · cos[(3θ−θ)/2] − cos2θ = 0
or, 2cos2θ · cosθ − cos2θ = 0
or, cos2θ(2cosθ − 1) = 0
Either,
cos2θ = 0
cos2θ = cos90°
2θ = 90°
θ = 45°.
OR,
2cosθ − 1 = 0
cosθ = 1/2
cosθ = cos60°
∴ θ = 60°.
(d) sin4n + sin2n = cos n
sin4n + sin2n − cosn = 0
or, 2sin[(4n+2n)/2] · cos[(4n−2n)/2] − cosn = 0
or, 2sin3n · cosn − cosn = 0
or, cosn(2sin3n − 1) = 0
Either,
2sin3n − 1 = 0
sin3n = 1/2
sin3n = sin30°
3n = 30°
n = 10°.
OR,
cosn = 0
Again,
sin3n = sin(180° − 30°)
3n = 150°
n = 50°.
OR,
cosn = 0
cosn = cos90°
∴ n = 90°.
(e) [First two terms partly unclear in source] + sin9A = 0
2sin9A · cos3A + sin9A = 0
or, sin9A(2cos3A + 1) = 0
Either,
sin9A = 0
sin9A = sin0°
9A = 0°
A = 0°
Again, sin9A = sin(180° − 0°)
9A = 180°
A = 20°.
OR,
cos3A = −1/2
cos3A = cos(180° − 60°)
3A = 120°
A = 40°
Again, cos3A = cos(180° + 60°)
3A = 240°
A = 80°.
(f) cos7θ − cosθ = −sin4θ
−2sin[(7θ+θ)/2] · sin[(7θ−θ)/2] = −sin4θ
or, −2sin4θ · sin3θ + sin4θ = 0
or, sin4θ(−2sin3θ + 1) = 0
Either,
sin4θ = 0
sin4θ = sin0°
4θ = 0°
θ = 0°
Again, sin4θ = sin(180° − 0°)
4θ = 180°
θ = 45°.
OR,
−2sin3θ + 1 = 0
sin3θ = 1/2
sin3θ = sin30°
3θ = 30°
θ = 10°
Again, sin3θ = sin(180° − 30°)
sin3θ = sin150°
3θ = 150°
θ = 50°.
(g) cos n + cos2n + cos3n = 0
2cos[(n+3n)/2] · cos[(n−3n)/2] + cos2n = 0
or, 2cos2n · cosn + cos2n = 0
or, cos2n(2cosn + 1) = 0.
Either,
cos2n = 0
cos2n = cos90°
2n = 90°
n = 45°
Again, cos2n = cos(360° − 0°)
2n = 360°
n = 180°. [as written in source]
OR,
cosn = −1/2
cosn = cos(180° − 60°)
n = 120°
Again, cosn = cos(180° + 60°)
n = 240°.
(h) √3/sin2A + 1/cos2A = 4 (0° ≤ A ≤ [Unreadable in source])
sin60°/(cos60°·sin2A) + 1/cos2A = 4
or, [sin60°·cos2A + cos60°·sin2A] / [cos60°·sin2A·cos2A] = 4
or, sin(60° + 2A) / [(1/2)sin2A·cos2A] = 4
or, sin(60° + 2A) / [sin2A·cos2A] = 2.
or, sin(60° + 2A) = 2sin2A·cos2A
or, sin(60° + 2A) = sin4A
or, 60° + 2A = 4A
or, 2A = 60°
or, A = 60°/2
∴ A = 30°.
4) If 2sin n · sin y = 1/2 and cot n + cot y = 2. Find the value of n + y.
→ cot n + cot y = 2
cosn/sinn + cosy/siny = 2
or, [cosn·siny + cosy·sinn] / [sinn·siny] = 2
or, sin(n + y) = 2sinn·siny
or, sin(n + y) = 1/2.
or, sin(n + y) = sin30°
or, n + y = 30°.
5) Solve: sin 2n = 3 tan n · cos 2n
2cosn·sinn = 3 × (sinn/cosn) × (2cos²n − 1)
or, 2cosn = (6cos²n − 3) / cosn
or, 2cos²n = 6cos²n − 3
or, 4cos²n − 3 = 0
or, cosn = ±√(3/4)
∴ cosn = √3/2.
Taking +ve:
cosn = √3/2
cosn = cos30°
n = 30°
Taking −ve:
Again, cosn = cos(360° − 30°)
cosn = cos330°
∴ n = 330°
Taking −ve:
cosn = cos(180° − 30°)
n = 150°
Again,
cosn = cos(180° + 30°)
n = 210°.
6) Solve: tanθ + tan2θ + √3 tanθ·tan2θ = √3
tanθ + tan2θ = √3 − √3 tanθ·tan2θ
or, tanθ + tan2θ = √3(1 − tanθ·tan2θ)
or, [tanθ + tan2θ] / [1 − tanθ·tan2θ] = √3
or, tan(θ + 2θ) = √3
or, tan3θ = tan60°
or, 3θ = 60°
∴ θ = 20°.
Again,
tan3θ = tan(180° + 60°)
or, 3θ = 240°
∴ θ = 80°.
Exercise 9.6
Starts again on physical PDF page 48. Source page order preserved.
Exercise 9.6
2) Find the values of n, y and θ.
(a)
→ In ΔABC
tan45° = n/20
or, 1 = n/20
∴ n = 20 m.
(b)
In ΔPQR
sin30° = y/16
or, 1/2 = y/16.
or, 2y = 16
or, y = 16/2
∴ y = 8 m.
(c)
cos θ = 10/20
or, cos θ = 1/2
or, cos θ = cos60°
∴ θ = 60°.
B) Short type questions
1) Find the value of n and y from given figures.
(a)
→ In ΔABC.
tan30° = n/(y + 10)
or, 1/√3 = n/(y + 10)
or, √3n = y + 10 …(i)
In ΔABD.
tan45° = n/y
or, 1 = n/y
∴ n = y …(ii)
From eqn. (i)
√3 y = y + 10
or, 1.73y − y = 10
or, 0.73y = 10.
or, y = 10 / 0.73
∴ y = 13.6986 m.
∴ y = n = 13.698 m.
(b)
→ In ΔABC
tan60° = 100/n
or, √3 = 100/n
or, n = 100/√3
∴ n = 57.7350 m.
In ΔABD.
tan22° = 100/y
or, y = 100 / 0.4040
∴ y = 247.52 m.
(c)
→ In ΔDBC.
tan45° = n/y
or, 1 = n/y
or, n = y …(i)
In ΔABC.
tan60° = (12 + n)/y
or, √3 = (12 + n)/y
or, √3n = 12 + n
or, √3n − n = 12.
or, n = 12 / 0.732050
∴ n = 16.3923 m.
(d)
→ In ΔAFC.
tan30° = AF/y
or, 1/√3 = AF/y
or, y = √3 AF …(i)
In ΔABD.
tan60° = (AF + 90)/y
or, √3 = (AF + 90)/(√3 AF)
or, 3AF = AF + 90
or, 3AF − AF = 90
or, 2AF = 90
or, AF = 45 m.
2) In the given figure, D is the mid-point of BC. Find the value of cot y / cot x.
→ In ΔABD.
cot x = AB/BD …(i)
In ΔABC.
cot y = AB/BC
or, cot y = AB/(BD + DC)
or, cot y = AB/(2BD) …(ii)
Now,
cot y / cot x = (AB/2BD) / (AB/BD)
or, cot y / cot x = AB/(2BD) × BD/AB
∴ cot y / cot x = 1/2.
C) Long type questions
1) The angle of elevation of the top of a tower from a point was observed to be 45°. On walking 30 m away from that point, it was found to be 30°. Find the height of the tower.
Let AB be the height of tower. C be a point where angle of elevation to the top of tower be 45° and D be the point 30 m away from C where angle of elevation be 30°.
→ In ΔABC.
tan45° = AB/BC
or, 1 = AB/BC
AB = BC …(i)
In ΔABD.
tan30° = AB/DB
or, 1/√3 = AB/(30 + AB)
or, √3AB = 30 + AB
or, 1.73AB − AB = 30
or, 0.73AB = 30
or, AB = 30/0.73
∴ AB = 41.09 m.
2(a) From the top of 21 m high cliff, the angles of depression of the top and the bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower.
→ In ΔADE.
tan45° = p/b
or, √3 = 21/DE [as written in source]
or, DE = 21/√3
∴ DE = 7√3.
In ΔABC.
tan45° = p/b
or, 1 = n/(7√3)
or, n = 7√3.
Now, CD = AE − AB
= 21 − 7√3
= 8.8756 m.
2(b) If the angles of depression and elevation of the top of a pole 25 m high from the top and bottom of a tower are 60° and 30° respectively. Find the height of the tower.
→ In ΔEDB.
tan30° = p/b
or, 1/√3 = 25/DB
or, DB = 25√3 m.
In ΔACE.
tan60° = p/b
or, √3 = h/(25√3)
or, h = √3 × 25√3
∴ h = 75.
height of tower = 75 + 25
= 100 m.
2(c) The angles of depression and elevation of the top of a tower 50 meter high from the top and bottom of a second tower are 45° and 30° respectively. Find the height of the second tower.
→ In ΔABC.
tan30° = 50/AC
or, 1/√3 = 50/EB
∴ EB = 50√3 m.
In ΔBED.
tan45° = (n − 50)/EB
or, 1 = (n − 50)/(50√3)
or, n − 50 = 50√3
or, n = 50√3 + 50
∴ n = 136.6025 m.
3) From the top of a building 30 m high a man observes two persons sitting on the ground at the same side at angles of depression of 45° and 30°. How far apart are the two persons?
→ In ΔABC.
tan45° = 30/AC
or, 1 = 30/AC
∴ AC = 30 m.
In ΔABD.
tan30° = 30/(AC + x)
or, 1/√3 = 30/(30 + x)
or, 30 + x = 30√3
or, x = 30√3 − 30
∴ x = 21.9815 m.
4) A pole is surmounted on its top by a flagstaff. The angles of elevation of the top and the bottom of the flagstaff as observed from a point 30 meters away from the bottom of the pole are found to be 45° and 30° respectively. Find the height of the flagstaff.
→ In ΔBDC.
tan30° = p/b
or, 1/√3 = h/30
or, √3h = 30
or, h = 30/√3
∴ h = 10√3 m.
In ΔADC.
tan45° = p/b
or, 1 = (y + h)/30
or, 30 = y + 10√3
or, 30 − 10√3 = y
∴ y = 12.6799 m.
5) From a point on the horizontal plane, the angle of elevation of the top of a pillar standing on the same plane was observed and found to be 60° and the angle of elevation of a point 20 m below the top of the pillar was found to be 30°. Find the height of the pillar.
→ In ΔBCD.
tan30° = (h − 20)/CD
or, 1/√3 = (h − 20)/CD
or, CD = (h − 20)/(1/√3) …(i)
In ΔACD.
tan60° = h/CD
or, √3 = h / [(h − 20)√3]
or, 3 = h/(h − 20)
or, 3h − 60 = h.
or, 2h = 60
∴ h = 30 m.
6) The angle of elevation of the top of a tower as observed from two distances of 36 m and 16 m from the foot of the tower are found to be complementary. Find the height of tower.
→ In ΔABC.
tanθ = p/b
tanθ = n/16 …(i)
In ΔABD.
tan(90° − θ) = n/36
or, cotθ = n/36
or, 1/tanθ = n/36
or, 16/n = n/36.
or, 16/n = n/36
or, n² = 576
∴ n = 24 m.
7) Two pillars of the same height are situated on the road 100 m apart. From any point between the pillars, the angles of elevations observed to the top of pillars are 60° and 30°. What will be the height of the pillars?
→ In ΔABE.
tan60° = p/b
or, tan60° = AB/(100 − x)
or, √3 = AB/(100 − x) …(i)
In ΔDEC.
tan30° = p/b
or, tan30° = DC/x
or, 1/√3 = DC/x
or, x = √3 DC
or, x = √3 AB …(ii)
From (i) and (ii),
√3/1 = AB/(100 − √3AB)
or, AB = √3(100 − √3AB)
or, AB = 100√3 − 3AB
or, AB + 3AB = 100√3
or, AB = 100√3/4
∴ AB = 25√3 m.
8) A poster hanging on a wall has a vertical height 3.66 m. From a point 5 m away from the wall on the same plane the angle of elevation of the bottom edge of the poster was found to be 45°. What will be the angle of elevation at the top edge of the poster as observed from the same point on the horizontal plane?
→ In ΔBCD.
tan45° = n/BD
or, 1 = n/5
∴ n = 5.
In ΔABD.
tanθ = (AC + n)/BD
or, tanθ = (3.66 + 5)/5
or, tanθ = 8.66/5
or, tanθ = tan−1(1.732)
∴ θ = 60°.
9)
AB is a vertical pole with its foot B on a leveled 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 20 m from the foot of the pole, find the height of the pole.
Discussion
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