A pole 6m high is fixed on the top of a tower. The angle of elevation

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Q) A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is 60 and the angle of depression of the point P from the top of the tower is 45. Find the height of the tower and the distance of point P from the foot of the tower. (use $\sqrt 3$ = 1.73)

Ans:

Step 1: Diagram for this question:

Let AB be the tower of H height and AC be the pole of 6 m on top of tower.

The distance between the Point P and the tower be D and angle of elevations be as shown in the image above.

Step 2: Calculating height H and distance D :

In Δ ABP, tan 45 = $\frac{AB}{BP}$

$1 = \frac{H}{D}$

∴ D = H …. (i)

Step 3: Find the height H:

Now in Δ CBP, tan 60 = $\frac{CB}{BP}$

$\sqrt 3 = \frac{H + 6}{D}$

$therefore H + 6 = D \sqrt 3$ …. (ii)

By substituting value of D from equation (i) in equation (ii), we get:

$H + 6 = H (\sqrt 3)$

$6 = H (\sqrt 3 - 1)$

$H = \frac{6}{\sqrt 3 - 1}$

$H = (\frac{6}{\sqrt 3 - 1})(\frac{\sqrt 3 + 1}{\sqrt 3 +1})$

$H = \frac{6 (\sqrt 3 + 1)}{(3 - 1)}$

$H = 3 (\sqrt 3 + 1)$ = 3 (1.73 + 1) = 3 (2.73) = 8.19 m

Therefore, the height of the tower is 8.19 m

Step 4: Find the distance D:

from equation (i), we get the distance D = H = 8.19 m

Therefore, the distance of the point P from the tower is 8.19 m.

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