Trigonometric Equations Question 137
Question: A vertical pole consists of two parts, the lower part being one third of the whole. At a point in the horizontal plane through the base of the pole and distance 20 meters from it, the upper part of the pole subtends an angle whose tangent is $ \frac{1}{2} $ . The possible heights of the pole are
[IIT 1964]
Options:
A) 20 m and $ 20\sqrt{3},m $
B) 20 m and 60 m
C) 16 m and 48 m
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \frac{H}{3}\cot \alpha =d $ and $ d=150\cot \varphi =60m $ or $ \frac{H}{3d}=\tan \alpha $ and $ \frac{H}{d}=\tan \beta $ $ \tan (\beta -\alpha )=\frac{1}{2}=\frac{\frac{H}{d}-\frac{H}{3d}}{1+\frac{H^{2}}{3d^{2}}} $
$ \Rightarrow $ $ 1+\frac{H^{2}}{3d^{2}}=\frac{4H}{3d} $
$ \Rightarrow $ $ H^{2}-4dH+3d^{2}=0 $
$ \Rightarrow $ $ H^{2}-80H+3,(400)=0 $
$ \Rightarrow $ $ H=20 $ or $ 60m $ .
BETA
