Trigonometric Equations Question 311
Question: The angles of elevation of the top of a tower from the top and bottom at a building of height a are $ 30^{o} $ and $ 45^{o} $ respectively. If the tower and the building stand at the same level, then the height of the tower is
[Karnataka CET 2000]
Options:
A) $ a\sqrt{3} $
B) $ \frac{a\sqrt{3}}{\sqrt{3}-1} $
C) $ \frac{a,(3+\sqrt{3})}{2} $
D) $ a,(\sqrt{3}-1) $
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta ABC,\tan 30^{o}=\frac{AC}{BC},\text{or }\frac{1}{\sqrt{3}}=\frac{x}{BC}, $ where $ AC=x, $ or $ BC=x\sqrt{3} $ and in $ \Delta ADE $ , $ \tan 45^{o}=\frac{a+x}{DE} $ or $ 1=\frac{a+x}{x\sqrt{3}} $ or $ x\sqrt{3}=a+x $ , $ x(\sqrt{3}-1)=a $ or $ x=\frac{a}{\sqrt{3}-1}. $ Therefore height of the tower, $ a+x=a+\frac{a}{\sqrt{3}-1} $ $ =a[ \frac{\sqrt{3}-1+1}{\sqrt{3}-1} ] $ $ =\frac{a\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} $ $ =\frac{a(3+\sqrt{3)}}{2}. $
BETA
