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 t

Question

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 $\sqrt{3} = 1.73$)

Accepted Answer

Step 1: Correct figure
Step 2: Let BC be the pole and AB be the tower of height '$h$' m.
$	an 45° = 1 = \frac{h}{x}$
$\Rightarrow h = x$ ... (i)
Step 3: $	an 60° = \sqrt{3} = \frac{h + 6}{x}$
$\Rightarrow h + 6 = x\sqrt{3}$ ... (ii)
Step 4: Solving (i) and (ii) to get
$h = 3(\sqrt{3} + 1) = 8.19$
Step 5: and $x = 8.19$
Therefore, the height of tower is $8.19$ m and the distance of point P from the foot of the tower is $8.19$ m

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