From the top of a hill 200 m high the angle of depression of the top and the bottom of a tower are observed to be 30° and 60°. The height of the tower is (in m);

Answer: Option C

Solution

Answer: Option C

Solution:

$$\begin{array}{rl}& AB=\text{hill}=200\text{ metre}\\ & CD=\text{tower}\\ & \text{In }\Delta APC\\ & \tan 30^{\circ }=\frac{AP}{PC}\\ & \frac{1}{\sqrt{3}}=\frac{AP}{PC}\\ & \Rightarrow AP:PC=1:\sqrt{3}....\left(\text{i}\right)\\ & \text{In }\Delta ABD\\ & \tan 60^{\circ }=\frac{AB}{BD}\\ & \sqrt{3}=\frac{AB}{BD}\\ & \Rightarrow AB:BD=\sqrt{3}:1....\left(\text{ii}\right)\\ & PB=CD\text{ and }PC=BD\\ & \text{Now}\end{array}$$
\begin{array}{*{20}{c}} {AB}&:&{BD}&:&{AP} \\ {\sqrt 3 }&:&1&{}&{} \\ {}&{}&{\sqrt 3 }&:&1 \\ 3&:&{\sqrt 3 }&:&1 \end{array}
$$\begin{array}{rl}& CD=PB\Rightarrow AB-AP\\ & CD=3-1\\ & CD=2\text{ units}\\ & AB=3\text{ units}=200\text{ metre}\\ & CD=2\text{ units}=\frac{200}{3}\times 2=133\frac{1}{3}\text{metre}\end{array}$$