# 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 C
$P Q=100(3-\sqrt{3}) m$ In triangle ABQ, $\begin{array}{l}\tan60=\frac{\text{h}}{\text{BQ}}\\ \text{BQ}=\frac{\text{h}}{\sqrt{3}}\ldots\ldots\ldots\left(1\right)\\ \text{In Triangle }\triangle\text{ABP}\\ \tan45=\frac{\text{h}}{\text{BP}}\\ 1=\frac{\text{h}}{\text{BQ}+100(3-\sqrt{3})}\\ \text{BQ}+300-100\sqrt{3}=\text{h}\\ \text{BQ}=\text{h}-300+100\sqrt{3}\ …..\left(2\right)\\ \text{From Eq 1 & 2, We get;}\\ \frac{\text{h}}{\sqrt{3}}=h-300+100\sqrt{3}\\ \Rightarrow\text{h}=\sqrt{3}\text{h}-300\sqrt{3}+300\\ \Rightarrow\text{h}+300\sqrt{3}-300=\sqrt{3}\text{h}\\ \Rightarrow300\sqrt{3}-300=\sqrt{3}\text{h}-\text{h}\\ \Rightarrow300\left(\sqrt{3}-1\right)=\text{h}\left(\sqrt{3}-1\right)\\ \Rightarrow\text{h}=300\ \text{m}\end{array}$