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CBSE Class 10 Mathematics

NCERT Exemplar Solutions

Chapter 8 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS

EXERCISE 8.4

1. If cosec θ + cot θ = p, then prove that cos θ = .

Sol. cosec θ + cot θ = p [Given]

⇒

⇒

Squaring Both Sides, We get

⇒

⇒

⇒

⇒

[By using the rule of componendo and dividendo can be written as ]

So, by using componendo and dividendo, we have

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⇒

⇒ cos θ =

Hence, proved.

2. Prove that = tan θ + cot θ.

Sol. LHS

= cosec θ sec θ

RHS = tan θ + cot θ

= cosec θ · sec θ = LHS

Hence, proved.

3. The angle of elevation of the top of a tower from a certain point is 30°. If the observer

moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find

the height of the tower.

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Sol. Consider the height of the vertical tower (TW) = x m (let)

Ist position of observer at A makes angle of elevation at the top of tower is 30°.

Now, angle of elevation of the top of tower is increased by 15°

i.e., it becomes 30° + 15° = 45°.

In ΔTWB, we have

tan 45° =

⇒ 1 =

⇒ x = y (I)

Now, ΔTWA, we have

tan 30° =

⇒ [From (I)]

⇒ = 20 + x

⇒ = 20

⇒ = 20

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⇒

⇒

⇒ x = 10(1.732 + 1)

⇒ x = 10 × 2.732 = 27.32 m

Hence, the height of the tower is 27.32 m.

4. If 1 + sin2 θ = 3 sin θ cos θ, then prove that tan θ = 1, or

.

Sol. To solve an equation in θ, we have to change it into one trigonometric ratio.

Given trigonometric equation is

1 + sin2 θ = 3 sin θ cos θ

⇒ [Dividing by sin2 θ both sides]

⇒ cosec2 θ + 1 = 3 cot θ

⇒ 1 + cot2 θ + 1 – 3 cot θ = 0

⇒ cot2 θ – 3 cot θ + 2 = 0

⇒ cot2 θ – 2 cot θ – 1 cot θ + 2 = 0

⇒ cot θ(cot θ – 2) –1(cot θ – 2) = 0

⇒ (cot θ – 2) (cot θ – 1) = 0

⇒ cot θ – 2 = 0 or (cot θ – 1) = 0

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⇒ cot θ = 2 or cot θ = 1

⇒ tan θ = or tan θ = 1

Hence, tan θ = , or 1.

5. Given that sin θ + 2 cos θ = 1, then prove that 2 sin θ – cos θ = 2.

Sol. sin θ + 2 cos θ = 1 [Given]

On squaring both sides, we get

(sin θ)2 + (2 cos θ)2 + 2(sin θ) (2 cos θ) = 1

⇒ sin2 θ + 4 cos2 θ + 4 sin θ cos θ = 1

⇒ 1 – cos2 θ + 4 (1 – sin2 θ) + 4 sin θ cos θ = 1

⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4 sin θ cos θ = 1

⇒ –cos2 θ – 4 sin2 θ + 4 sin θ cos θ = –4

⇒ cos2 θ + 4 sin2 θ – 4 sin θ cos θ = 4

⇒ (cos θ)2 + (2 sin θ)2 – 2(cos θ) (2 sin θ) = 4

⇒ (2 sin θ – cos θ)2 = 22

Taking square root both sides, we have

2 sin θ – cos θ = 2

Hence, proved.

6. The angle of elevation of the top of a tower from two distant points s and t from its

foot are complementary. Prove that the height of tower is .

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Sol. Let the height of the vertical tower (TW) = x m

Points of observation A and B are at distances ‘t’ and ‘s’ from the foot of tower.

The angles of elevation of top of the tower from observation points A and B are (90° – θ) and

θ which are complementary.

In ΔTWB, we have

tan θ = (I)

Now, in ΔTWA, we have

tan (90° – θ) =

⇒ cot θ =

⇒ cot θ · tan θ = [Multiply eq (i) and (ii) ]

⇒

⇒

⇒ x2 = st

⇒ x =

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Hence, proved.

7. The shadow of a tower standing on a plane level is found to be 50 m. longer when

Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Sol. Let a tower TW of light x (let) is standing vertically upright on a level plane ABW. A and

B are two positions of observation when angle of elevation changes from 30° to 60°

respectively.

Let BW = y [Given]

AB = 50 m

In ΔTWB, we have

tan 60° =

⇒

⇒ x = (I)

Now, in ΔTWA, we have

tan 30° = (II)

⇒ [From (I)]

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⇒ 3y = y + 50

⇒ 3y – y = 50

⇒ 2y = 50

⇒ y = 25 (I)

Now, x =

⇒ x =

⇒ x = m

8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag

staff of height ‘h’. At a point on the plane, the angles of elevation on the bottom and top

of the flag staff are α and β, respectively. Prove that the height of the tower is

.

Sol. Let the height of vertical tower (TW) = x.

And, the height of flag staff (TF) = h (Given)

The angle of elevation at A on ground from the base and top of flag staff are α, β respectively.

Let AW = y

In ΔTWA, we have

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tan α =

⇒ y = (I)

Now, in ΔFWA, we have

tan β =

⇒ y tan β = x + h

⇒ = x+ h [From (I)]

⇒ x tan β = x tan α + h tan α

⇒ x(tab β – tan α) = h tan α

⇒

Hence, proved.

9. If tan θ + sec θ = l, then prove that sec θ = .

Sol. [Recall identity sec2 θ – tan2 θ = 1

and now change sec θ + tan θ to sec2 θ – tan2 θ by multiplying and dividing the given

expression to (sec θ – tan θ).

sec θ + tan θ = l [Given]

⇒ (sec θ + tan θ)

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⇒ [∵ 1 + tan2 θ = sec2 θ]

⇒

or sec θ – tan θ = (II)

Now, get sec θ by eliminating tan θ from (I) and (II).

It can be obtained by adding (I) and (II).

⇒ 2 sec θ =

⇒ 2 sec θ =

⇒ sec θ =

Hence, proved.

10. If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.

Sol. sin θ + cos θ = p

sec θ + cosec θ = q

[ IInd expression can be changed into sin θ, cos θ and eliminate trigonometric ratio from (I)

and (II)]

sec θ + cosec θ = q

⇒

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⇒

⇒ [Using (I)]

⇒ sin θ cos θ = (III)

sin θ + cos θ = p [From (I)]

⇒ (sin θ + cos θ)2 = p2 [Squaring both sides]

⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = p2

⇒ 1 + 2 · = p2 [∵ sin2 θ + cos2 θ = 1 and using (III)]

⇒ q + 2p = p2q

⇒ 2p = p2q – q

⇒ 2p = q(p2 – 1)

Hence, proved.

11. If a sin θ + b cos θ = c, then prove that a cos θ – b sin θ =

.

Sol. a sin θ + b cos θ = c [Given]

On squaring both sides, we get

(a sin θ)2 + (b cos θ)2 + 2(a sin θ) (b cos θ) = c2

⇒ a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ = c2

⇒ a2(1 – cos2 θ) + b2 (1 – sin2 θ) + 2 ab sin θ cos θ = c2

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⇒ a2 – a2 cos2 θ + b2 – b2 sin2 θ + 2ab sin θ cos θ = c2

⇒ –a2 cos2 θ – b2 sin2 θ + 2ab sin θ cos θ = c2 – a2 – b2

⇒ a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ = a2 + b2 – c2

⇒ (a cos θ)2 + (b sin θ)2 – 2(a cos θ) (b sin θ) = a2 + b2 – c2

⇒ (a cos θ – b sin θ)2 = a2 + b2 – c2

⇒ a cos θ – b sin θ = (Taking square root both sides)

Hence, a cos θ – b sin θ =

Hence, proved.

12. Prove that

Sol. Recall identity sec2 θ – tan2θ = 1

LHS

[∵ a2 – b2 = (a – b)(a + b)]

= sec θ – tan θ

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= RHS

Hence, proved.

13. The angle of elevation of the top of a tower 30 m high from the foot of another tower

in the same plane is 60°, and the angle of elevation of the top of second tower from the

foot of first tower is 30°. Find the distance between the two towers and also the height

of the other tower.

Sol. Two vertical tower TW = 30 m and ER = x m (let) are standing on a horizontal plane RW

= y(let). The angle of elevation from R to top 90 m high tower is 60° and the angle of elevation

of second tower from W is 30°.

In ΔERW,

tan 30° =

⇒

⇒ y = (I)

Now, In ΔTWR,

tan 60° = (I)

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⇒ [From (I)]

⇒ 3x = 30

⇒ x = 10 m

Now, y =

⇒ y =

⇒ y = 1.732 × 10

⇒ y = 17.32 m

Hence, the distance between the two towers is 17.32 m and the height of the second tower is

10 m.

14. From the top of a tower h m high, the angles of depression of two objects, which are

in line with the foot of the tower are α and β, (β > α). Find the distance between the two

objects.

Sol. Consider a vertical tower TW = h m. Two objects A and B are x m apart in the line joining

A, B and W and BW = y(let).

The angle of depression from top a tower to objects A and B are α and β respectively.

In ΔTWB, we have

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tan β =

⇒ y = (I)

Now, in ΔTWA,

tan α =

⇒ tan α (x + y) = h

⇒ tan α = h [From (I)]

⇒ x tan α + = h

⇒ x tan α = h –

⇒ x tan α =

⇒ x =

⇒ x =

⇒ x =

⇒ x = h[cot α – cot β]

Hence, the distance between the two objects is h(cot α – cot β) m.

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15. A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is

pulled away from the wall through a distance p, so that its upper end slides a distance q

down the wall and then the ladder makes an angle β with horizontal. Show that

.

Sol. Consider a vertical wall WB. Two positions AW and LD of a ladder as shown in figure

such that LA = p, WD = q and LD = AW = z. Angle of inclination of ladder at two positions A

and L are α and β respectively.

Let AB = y and DB = x.

In ΔABW, we have

sin α =

and cos α =

In ΔLBD, we have

Sin β =

and cos β =

Taking RHS =

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= LHS

Hence, proved.

16. The angle of elevation of the top of a vertical tower from a point on the ground is

60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find

the height of the tower.

Sol. Let the height of the vertical tower TW = x m.

It stands on a horizontal plane AW = y.

Also BC = y.

Observation point B is 10 m above the first observation point A.

The angles of elevation from point of observations A and B are 60° and 45° respectively.

TC = x – 10

In right angled triangle TBC, we have

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tan 45° =

⇒ 1 =

⇒ y = x – 10 (I)

Now, in ΔTAW,

tan 60° =

⇒ [From (I)]

⇒ = x

⇒

⇒

⇒ x =

⇒ x =

⇒ x = 23.66 m

= 100 × 2.366

Hence, the height of the tower = 23.66 m.

17. A window of a house is h m above the ground. From the window, the angles of

elevation and depression of the top and the bottom of another house situated on the

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opposite side of the lane are found to be α and β respectively. Prove that the height of

the other house is h(1 + tan α cot β) m.

Sol. Window W, h m above the ground point A, another house HS = x(m), AS = y m away

from observation window, AS = WN = y (let), NS = h, HN = (x – h).

Angle of elevation and depression of top and bottom of house HS from window W are α, β

respectively.

In right angled ΔWNS,

tan β =

⇒ y = (I)

Now, in right angled ΔHNW,

⇒ tan α =

⇒ y tan α = x – m h

⇒ = x – h [From (I)]

⇒ h tan α = x tan β – h tan β

⇒ x tan β = h tan α + h tan β

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⇒ x =

⇒ x =

⇒ x = h[tan α · cot β + 1]

Hence, the height of the house on the other side of the observer is h[1 + tan α · cot β] m.

18. The lower window of a house is at a height of 2 m above the ground and its upper

window is 4 m vertically above the lower window. At certain instant the angles of

elevation of a balloon from these windows are observed to be 60° and 30°, respectively.

Find the height of the balloon above the ground.

Sol. Let B be a balloon at a height GB = x m.

Let W1 be the window, which is 2 m above the ground H.

∴ W1 H = 2

⇒ AG = 2 m

Let W2 be the second window, which is the 4 m above the window W1.

∴ W2W1 = AC = 4 m

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The angles of elevation of balloon B from W1, W2 are 60° and 30° respectively.

BA = (x – 2) m

BC = x – 2 – 4 = (x – 6) m

In right angled ΔW2CB, we have

tan 30° =

⇒

⇒ y =

Now, in right angled ΔW1 AB, (I)

tan 60° =

⇒

⇒ = (x – 2) (II)

⇒ (x – 6) = x – 2 [From (I)]

⇒ 3x – 18 = x – 2

⇒ 3x – x = 18 – 2

⇒ 2x = 16

⇒ x = 8 m

Hence, the height of the balloon above the ground is 8 m.

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