Options:
A) $ \sqrt{\frac{a}{b}} $
B) $ \sqrt{a+b} $
C) $ \sqrt{ab} $
D) $ \sqrt{a-b} $
Show Answer
Answer:
Correct Answer: C
Solution:
- In the above figure, there are two elevation angles $ 60{}^\circ $ and $ 30{}^\circ $ of a tower.
Let the height of the tower he h and distance points are a and $ b(a>b). $
Now, in $ \Delta ACB $
$ \tan 60{}^\circ =\frac{AB}{BC}=\frac{h}{b} $
$ \Rightarrow $ $ \sqrt{3}=\frac{h}{b} $
$ h=b\sqrt{3} $
(i)
Now, again in $ \Delta ADB $
$ \tan 30{}^\circ =\frac{AB}{BD}=\frac{h}{a} $
$ \Rightarrow $ $ \frac{1}{\sqrt{3}}=\frac{h}{a} $
$ \Rightarrow $ $ h=\frac{a}{\sqrt{3}} $
(ii)
On multiplying Eqs. (i) and (ii), we get
$ h^{2}=(b\sqrt{3})\times ( \frac{a}{\sqrt{3}} ) $
$ \Rightarrow $ $ h^{2}=ab $
$ \therefore $ $ h=\sqrt{ab} $