Question

The angle of elevation of the top point P of the vertical tower PQ of height h from a point A on the ground is ${45}^{\circ }$ and from a point B, the angle of elevation is ${60}^{\circ }$, where B is a point at a distance d from the point A, measured along the line AB, which makes an angle ${30}^{\circ }$ with AQ. prove that d = $h(\sqrt{3}-1)$

Answer 1

In $\Delta PAQ$, we have

$\frac{AQ}{PQ}=cot{45}^{\circ }=1\Rightarrow \frac{AQ}{h}=1\Rightarrow AQ=h$

From right $\Delta AQP$, we have

$A{P}^2=A{Q}^2+P{Q}^2=(h^2+h^2)=2h^2\Rightarrow AP=\sqrt{2}h.$

$In\Delta PBH,\angle BPH={180}^{\circ }-({60}^{\circ }+{90}^{\circ })={30}^{\circ }$

$In\Delta PAQ,\angle APQ={180}^{\circ }-({40}^{\circ }+{90}^{\circ })={45}^{\circ }$

$\therefore \angle APB=(\angle APQ-\angle BPH)=({45}^{\circ }-{30}^{\circ })={15}^{\circ }$

In $\Delta APB,$ we have

$\angle PAB={15}^{\circ },\angle APB={15}^{\circ }and\angle ABP={180}^{\circ }-({15}^{\circ }+{15}^{\circ })={150}^{\circ }$

Using sine formula on $\Delta ABP,$ we have

$\frac{AB}{sin{15}^{\circ }}=\frac{AP}{sin{15}^{\circ }}$

$\Rightarrow \frac{d}{sin{15}^{\circ }}=\frac{\sqrt{2}h}{sin{30}^{\circ }}$

$\begin{array}{c}\Rightarrow d=2\sqrt{2}h.sin{15}^{\circ }\\ \Rightarrow d=2\sqrt{2}h.\frac{(\sqrt{3}-1)}{2\sqrt{2}}\\ =(\sqrt{3}-1)h.\\ \text{Hence},d=(\sqrt{3}-1)h.\end{array}\left[\begin{array}{c}\because sin{15}^{\circ }=(sin{45}^{\circ }-{30}^{\circ })\\ =sin{45}^{\circ }cos{30}^{\circ }-cos{45}^{\circ }sin{30}^{\circ }\\ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{(\sqrt{3}-1)}{2\sqrt{2}}\end{array}\right]$