The angle of elevation of the top of a tower from a certain point is 30°. If the observed moves 20 m towards the tower, the angle of elevation the angle of elevation of top of the tower increases by 15°. The height of the tower is

Answer: Option C

Solution

Answer: Option C

Solution:

Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\begin{array}{rl}& \frac{AB}{AC}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}\\ & \Rightarrow AC=AB\times \sqrt{3}=h\sqrt{3}\text{And,}\\ & \Rightarrow \frac{AB}{AD}=\tan 45^{\circ }=1\\ & \Rightarrow AD=AB=h\\ & CD=20\\ & \Rightarrow \left(AC-AD\right)=20\\ & \Rightarrow h\sqrt{3}-h=20\\ & \therefore h=\frac{20}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)}\\ & =10\left(\sqrt{3}+1\right)\text{m}\\ & =\left(10\times 2.73\right)\text{m}\\ & =27.3\text{m}\end{array}$$