From the top of a tower, the angles of depression of two objects P and Q (situated on the ground on the same side of the tower) separated at a distance of $100(3-\sqrt{3})$ m are 45° and 60 ° respectively. The height of the tower is.
A. 200 m B. 250 m C. 300 m D. None of these Answer: Option CShow Answer
Solution(By Apex Team)
$P Q=100(3-\sqrt{3}) m$
In triangle ABQ,
$\begin{array}{l}\tan60=\frac{\text{h}}{\text{BQ}}\\
\text{BQ}=\frac{\text{h}}{\sqrt{3}}\ldots\ldots\ldots\left(1\right)\\
\text{In Triangle }\triangle\text{ABP}\\
\tan45=\frac{\text{h}}{\text{BP}}\\
1=\frac{\text{h}}{\text{BQ}+100(3-\sqrt{3})}\\
\text{BQ}+300-100\sqrt{3}=\text{h}\\
\text{BQ}=\text{h}-300+100\sqrt{3}\ …..\left(2\right)\\
\text{From Eq 1 & 2, We get;}\\
\frac{\text{h}}{\sqrt{3}}=h-300+100\sqrt{3}\\
\Rightarrow\text{h}=\sqrt{3}\text{h}-300\sqrt{3}+300\\
\Rightarrow\text{h}+300\sqrt{3}-300=\sqrt{3}\text{h}\\
\Rightarrow300\sqrt{3}-300=\sqrt{3}\text{h}-\text{h}\\
\Rightarrow300\left(\sqrt{3}-1\right)=\text{h}\left(\sqrt{3}-1\right)\\
\Rightarrow\text{h}=300\ \text{m}\end{array}$
