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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

In right Î”ADE, we have:

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‡’ h^{2} = a Ã— b Ã— (cot Î¸ Ã— tan Î¸) [âˆµ cot Î¸ Ã— tan Î¸ = 1]

â‡’ h^{2} = 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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