A tower PQ stands on a horizontal ground with base Q on the ground.

Question

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR=15$m. If from a point $A$ on the ground the angle of elevation of $R$ is $60^\circ$ and the part $PR$ of the tower subtends an angle of $15^\circ$ at $A$, then the height of the tower is:

Accepted Answer

For $$\Delta$$AQR,

$$\tan 60^\circ  = {{15} \over x}$$ ....... (1)

From $$\Delta$$AQP,

$$\tan 75^\circ  = {h \over x}$$

$$ \Rightarrow \left( {2 + \sqrt 3 } \right) = {h \over x}$$ [$$\because$$ $$\tan 75^\circ  = 2 + \sqrt 3 $$]

$$ \Rightarrow h = \left( {2 + \sqrt 3 } \right)x$$

$$ = \left( {2 + \sqrt 3 } \right){{15} \over {\sqrt 3 }}$$ [From (1)]

$$ = \left( {2 + \sqrt 3 } \right) \times {{15\sqrt 3 } \over 3}$$

$$ = \left( {2 + \sqrt 3 } \right) \times 5\sqrt 3 $$

$$ = 5\left( {2\sqrt 3  + 3} \right)$$ m

Solutions · 1