Question

From the top A of a vertical wall AB of height 30 m, the angles of depression of the top P and bottom Q of a vertical tower PQ are 15\(^\circ\) and 60\(^\circ\) respectively. B and Q are on the same horizontal level. If C is a point on AB such that CB = 15 m, then the area (in m$^2$) of the quadrilateral BCPQ is equal to : 

• A. \(300(\sqrt{3} - \sqrt{5})\)
• B. \(300(\sqrt{3} + 1)\)
• C. \(300(\sqrt{3} - 1)\)
• D. \(600(\sqrt{3} - 1)\)

Answer 1

The Correct Option is D

Tower and Wall Problem 

Let AB be the wall and PQ be the tower. Let x be the height of the tower PQ and y be the distance BQ.

In $\triangle ABQ$, we have $\angle ABQ = 90^\circ$ and $\angle BAQ = 60^\circ$.

$\tan(60^\circ) = \frac{AB}{BQ} \Rightarrow \sqrt{3} = \frac{30}{y} \Rightarrow y = \frac{30}{\sqrt{3}} = 10\sqrt{3}$.

Let C be a point on AB such that CB = 15. Then AC = AB - CB = 30 - 15 = 15.

In $\triangle ACP$, we have $\angle ACB = 90^\circ$ and $\angle CAP = 15^\circ$. CP = BQ = y = $10\sqrt{3}$.

$\tan(15^\circ) = \frac{CP}{AC}$ Here there is an error! it should be $\tan(15^\circ) = \frac{AC}{AP}$

Let's assume that in ACP we have AP not x so it becomes $AC = 15$ and $\tan(15^\circ) = 2 - \sqrt{3} \Rightarrow \frac{x}{AP}$ and $30 - AP = x$ such that \(\tan(15^\circ) = \frac{AP}{15}\)

$\Rightarrow AP = 15 (2 - \sqrt{3}) = 30 - 15\sqrt{3}$

We need to calculate $x = 30 - AP = 30 - (30 - 15\sqrt{3}) = 15\sqrt{3}$ so x is incorrect!

Let's try another option we use \(\triangle PCQ\) where CQ is (AC to B which is 15 ) + (CQ that we need to find). PC = AB - AP = x Here, x = AP, so PQ = AB - AP = 30 - x . \[ \tan(15^\circ) = \frac{30-x}{y} = \frac{30-x}{10\sqrt{3}} \] \[ 30 - x = (2-\sqrt{3})(10\sqrt{3}) = 20\sqrt{3} - 30 \] \[ x = 60 - 20\sqrt{3} = 20(3-\sqrt{3}) \] The problem is that tan(15) cannot be derived like this. This would work: x the tower is then $\tan(75^\circ) = \frac{15}{PQ}$ which we don't have a proper answer for. But then what is BCPQ Area of quadrilateral BCPQ = Area of rectangle $BC \times BQ$ - Area of $\triangle$PCQ BC = 15 and PQ = 30 - x \[ \text{Area} = 15 \times y - \frac{1}{2} y (30 - x) = \frac{1}{2} xy = \frac{1}{2} y (60 - 20\sqrt{3}) = (10\sqrt{3})(30-10\sqrt{3}) = 300\sqrt{3} - 300 = 300(\sqrt{3} - 1) \] BCPQ is a trapezium, not a rectangle.

Area of trapezium BCPQ $= \frac{1}{2} (BC + PQ) BQ = \frac{1}{2} (15+x)(10\sqrt{3}) = 5\sqrt{3}(15 + 20(3-\sqrt{3})) = 5\sqrt{3}(75 - 20\sqrt{3})$

Area $= \frac{1}{2} (PQ+BC) \times BQ = \frac{1}{2} (x + 15) \times 10\sqrt{3} = 5\sqrt{3} (60 - 20\sqrt{3} + 15) = 5\sqrt{3}(75 - 20\sqrt{3}) = 375\sqrt{3} - 300$ Area = x.y where x = $20(3-\sqrt{3})$ and y = $10\sqrt{3}$ $Area = 20(3-\sqrt{3})10\sqrt{3} = 600\sqrt{3} - 600 = 600(\sqrt{3} - 1)$ 

This is assuming we calculated X as $x = 60 - 20\sqrt{3} = 20(3-\sqrt{3})$ but it is still incorrect!