Trigonometric Equations Question 60
Question: $ ABCD $ is a rectangular field. A vertical lamp post of height 12m stands at the corner A. If the angle of elevation of its top from B is $ 60^{o} $ and from C is $ 45^{o} $ , then the area of the field is
[Kerala (Engg.) 2005]
Options:
A) $ 48\sqrt{2}sq.m $
B) $ 48\sqrt{3}sq.m $
C) $ 48sq.m $
D) $ 12\sqrt{2}sq.m $
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Answer:
Correct Answer: A
Solution:
- Let AE is a vertical lamp-post. Given, $ AE=12 $ m $ \tan 45^{o}=\frac{AE}{AC} $ $ AC=AE=12m $ $ \tan 60^{o}=\frac{AE}{AB} $ $ AB=\frac{AE}{\sqrt{3}}=4\sqrt{3} $ $ BC=\sqrt{AC^{2}-AB^{2}} $ $ =\sqrt{144-48} $ $ =\sqrt{96} $ $ =4\sqrt{6} $ Area = $ AB\times BC=4\sqrt{3}\times 4\sqrt{6} $ = $ 48\sqrt{2} $ sq. cm.
BETA
