A and B are standing on ground 50 meters apart. The angles of elevation for these two to the top of a tree are 60° and 30°. What is height of the tree?

Answer: Option C

Solution

Answer: Option C

Solution:

$$\begin{array}{rl}& \text{In}\Delta PBQ,\tan 60^{\circ }=\frac{PQ}{BQ}\\ & \therefore BQ=\frac{PQ}{\sqrt{3}}\\ & \text{In}\Delta PAQ,\tan 30^{\circ }=\frac{PQ}{AQ}\\ & \therefore \frac{1}{\sqrt{3}}=\frac{PQ}{50+BQ}\\ & \therefore PQ=\frac{50+BQ}{\sqrt{3}}\\ & =\frac{50+\frac{PQ}{\sqrt{3}}}{\sqrt{3}}\\ & =\frac{50\sqrt{3}+PQ}{\sqrt{3}\times \sqrt{3}}\\ & \therefore 3PQ=50\sqrt{3}+PQ\end{array}$$
$$\therefore PQ=25\sqrt{3}\text{m}=$$     Height of tree