Question

10. Find the
1.782)
3. Prom a point on the ground which is 120
from the foot of the unfinished tower, the ang
of elevation of the pop of the tower is found
30° find how much height of tower hans
increased so that the angle of elevation at sa
point became 60%​

Answer 1

Step-by-step explanation:

Let PA be the unfinished tower.

Let B be the point of observation i.e. 120 m away from the base of the tower.

Now, AB = 120m

Let, ∠ABP=45°

Let h m be the height by which the unfinished tower be raised such that its angle of elevation of the top from the same point becomes 60°.

Let CA = h &∠ABC=60°

In triangle ABP,tan45°=Ab

Now, AB = 120m

Let, ∠ABP=45°

Let h m be the height by which the unfinished tower be raised such that its angle of elevation of the top from the same point becomes 60°.

Let CA = h &∠ABC=60°

In triangle ABP,

tan45°=

AB

PA

⇒1=

120

PA

⇒PA=120M

Now, in triangle ABC,

tan60°=

AB

CA

3

=

120

120+h

h+120=120

3

h=120(

3

−1)

h=120(1.732−1)→(as

3

=1.732)

h=120×0.732

h=87.84m

Answer 2

Answer:

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