Trigonometric Equations Question 45
Question: A vertical pole (more than 100 m high) consists of two portions, the lower being one-third of the whole. If the upper portion subtends an angle $ {{\tan }^{-1}}\frac{1}{2} $ at a point in a horizontal plane through the foot of the pole and distance 40 ft from it, then the height of the pole is
[AMU 1981]
Options:
A) 100 ft
B) 120 ft
C) 150 ft
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Obviously, from figure $ \tan \alpha =\frac{h/3}{40}=\frac{h}{120} $ …..(i) $ \tan \beta =\frac{h}{40}=\frac{3h}{120} $ …..(ii) Therefore $ \tan \theta =\tan (\beta -\alpha ) $
$ \Rightarrow $ $ \frac{1}{2}=\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3h^{2}}{14400}}\Rightarrow h=120,,40 $ But $ h=40 $ cannot be taken according to the condition, therefore $ h=120ft $ .
BETA
