The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on …

Question

The angle of elevation of the top $P$ of a vertical tower $PQ$ of height 10 from a point $A$ on the horizontal ground is $45^\circ$. Let $R$ be a point on $AQ$ and from a point $B$, vertically above $R$, the angle of elevation of $P$ is $60^\circ$. If $\angle BAQ=30^\circ$, $AB=d$ and the area of the trapezium $PQRB$ is $\alpha$, then the ordered pair $(d,\alpha)$ is:

Accepted Answer

Let $BR = x$

$${x \over d} = {1 \over 2} \Rightarrow x = {d \over 2}$$

$${{10 - x} \over {10 - x\sqrt 3 }} = \sqrt 3  \Rightarrow 10 - x = 10\sqrt 3  - 3x$$

$$2x = 10(\sqrt 3  - 1)$$

$$x = 5(\sqrt 3  - 1)$$

$$d = 2x = 10(\sqrt 3  - 1)$$

$$\alpha  = {1 \over 2}(x + 10)(10 - x\sqrt 3 )=$$ Area (PQRB)

$$ = {1 \over 2}\left( {5\sqrt 3  - 5 + 10} \right)\left( {10 - 5\sqrt 3 (\sqrt 3  - 1)} \right)$$

$$ = {1 \over 2}\left( {5\sqrt 3  + 5} \right)\left( {10 - 15 + 5\sqrt 3 } \right) - {1 \over 2}(75 - 25) = 25$$

Solutions · 1