Question

From the top of a tower, $80\text{m}$ high, the angles of depression of two points $P$ and $Q$ in the same vertical plane with the tower are ${45}^{{}^{\circ }}$ and ${75}^{{}^{\circ }}$ respectively, find the value of $PQ.$

[Use $\tan {75}^{\circ }=2+\sqrt{3}$]

• A. $80(\sqrt{3}+1)\text{m}$
• B. $80(\sqrt{3}-1)\text{m}$
• C. $160(\sqrt{3}+1)\text{m}$
• D. $160(\sqrt{3}-1)\text{m}$

Answer 1

The correct option is B $80(\sqrt{3}-1)\text{m}$

Let the distance of point $Q$ from the base of the tower be $y$ and distance of point $P$ from $Q$ be $x.$

In $△QSR,$

$\tan {75}^{\circ }=\frac{SR}{y}$

$\Rightarrow (2+\sqrt{3})=\frac{80}{y}$

$\Rightarrow y=\frac{80}{(2+\sqrt{3})}$

Now, in $△PSR,$

$\begin{array}{cc}\tan {45}^{\circ }& =\frac{SR}{x+y}\\ =1& =\frac{80}{x+y}\\ x+y& =80\\ x+\frac{80}{2+\sqrt{3}}& =80\\ x& =80-\frac{80}{2+\sqrt{3}}\\ & =\frac{80\left(\sqrt{3}+1\right)}{\sqrt{3}+2}\\ & =80\left(\sqrt{3}-1\right)\text{m}\end{array}$

Hence, $PQ$ is equal to $80(\sqrt{3}-1)\text{m}.$