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# Ex 9.2 NCERT Solutions- Some Applications of Trigonometry Class 10 Notes | EduRev

## Class 10 : Ex 9.2 NCERT Solutions- Some Applications of Trigonometry Class 10 Notes | EduRev

The document Ex 9.2 NCERT Solutions- Some Applications of Trigonometry Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 9.2

Q 1. The angles of depression of the top and the bottom of a building 50 m high as observed from the top of a tower are 30Â° and 60Â° respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Sol. In the figure

Let AB = 50 m be the building.
Let CE be the tower such that CE = (50 + x) m

â‡’         ...(1)

In right Î”ACE, we have:

â‡’      ...(2)

From (1) and (2), we get

â‡’
â‡’ 3x âˆ’ x =50 â‡’ x = 25

âˆ´ Height of the tower = 50 + x
= 50 + 25
= 75 m

Now from (1), BC = âˆš3 Ã— x
= âˆš3 Ã— 25 m
= 1.732 Ã— 25 m
= 43.25 m

i.e., The horizontal distance between the building and the tower = 43.25 m.

Q 2. The angle of elevation of the top of a tower as observed from a point on the ground is â€˜Î± â€™ and on moving â€˜aâ€™ metres towards the tower, the angle of elevation is â€˜Î²â€™. Prove that the height of the tower is

Sol. In the figure, let the tower be represented by AB.
âˆ´ In right Î” ABC, we have:

â‡’ x tan Î² = h

â‡’

Now, in right Î”ABD, we have:

Q 3. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 m. From a point on the plane the angles of elevation of the bottom and top of the flag staff are respectively 30Â° and 60Â°. Find the height of the tower.

Sol. Let in the figure, BC be the tower such that

BC = y metres.

CD be the flag staff such that
CD = 5m
â‡’ BD =(y + 5) m.

In right Î” ABC, we have:

â‡’      ...(1)

In right Î” ABD, we have:

â‡’

âˆ´

â‡’ y + 5 = 3 y
â‡’ 3y âˆ’ y = 5 â‡’ y = 5/2  = 2.5 m

âˆ´ The height of the tower = 2.5 m.

Q 4. The length of the shadow of a tower standing on level plane is found to be 20 m longer when the sunâ€™s altitude is 30Â° than when it was 60Â°. Find the height of the tower.

Sol. In the figure, let CD be the tower such that
CD = h metres
Also BC = x metres

â‡’

In right Î” ACD, we have:

â‡’

â‡’

Thus, the height of the tower = 17.32 m.

Q 5. From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30Â° and 60Â° respectively. Find the height of the pillar and its distance from the hill.

Sol. In the figure, let AD is the hill such that
AD = 200 m and CE is the pillar.

âˆ´

â‡’

â‡’

â‡’ Distance between pillar and hill = 115.33 m

Now,      [âˆµ DE = BC]

In right âˆ† ABC, we have:

â‡’

âˆ´ Height of the pillar

CE = AD âˆ’ AB    [âˆ´ CE = BD]
= 200 âˆ’ 66.67 m
= 133.33 m

Q 6. The angles of elevation of the top of a tower from two points on the ground at distances a and b units from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is  units.

Sol. In the figure, AB is the tower, such that:

AB = h
BD = b
BC = a

In right Î” ABD, we have

â‡’
â‡’  h = b cot Î¸     ...(1)

In right Î” ABC, we have

â‡’      ..(2)

Multiplying (1) and (2), we get
h Ã— h = b cot Î¸ Ã— a tan Î¸
â‡’ h2 = a Ã— b Ã— (cot Î¸ Ã— tan Î¸)    [âˆµ cot Î¸ Ã— tan Î¸ = 1]
â‡’ h2 = a b
â‡’

Q  7. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point â€˜Aâ€™ on the ground is 60Â° and the angle of depression of the point â€˜Aâ€™ from the top of the tower is 45Â°. Find the height of the tower.

Sol. In the figure, let BC be the tower and CD be the pole.
Let BC = x metres and AB = y metres In right âˆ† ABC, we get

â‡’ BC = AB â‡’ y = x ... (1)

In right Î” ABD, we have:

â‡’
â‡’
â‡’
âˆ´        [âˆµ x = y from (1)]
â‡’
â‡’

Thus, the height of the tower = 6.83 m

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