Vector-Algebra Question 411
Question: The upper $ \frac{3}{4} $ th portion of a vertical pole subtends an angle $ {{\tan }^{-1}}\frac{3}{5} $ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is
Options:
A) 80 m
B) 20 m
C) 40 m
D) 60 m
Correct Answer: C
Show Answer
Answer:
Solution:
$ \Rightarrow \tan \beta =\frac{\tan \theta -\tan \alpha }{1+\tan \theta .\tan \alpha } $ or $ \frac{3}{5}=\frac{\frac{h}{40}-\frac{h}{160}}{1+\frac{h}{40}.\frac{h}{160}} $
$ \Rightarrow h^{2}-200h+6400=0\Rightarrow h=40 $ or 160 metre
$ \therefore $ possible height = 40 metre
BETA
