Trigonometric Equations Question 324
Question: $ ABC $ is a right angled isosceles triangle with $ \angle B=90^{o} $ . If D is a point on $ AB $ so that $ \angle DCB=15^{o} $ and if $ AD=35cm $ , then $ CD= $
[Kerala (Engg.) 2005]
Options:
A) $ 35\sqrt{2} $ cm
B) $ 70\sqrt{2}cm $
C) $ \frac{35\sqrt{3}}{2}cm $
D) $ 35\sqrt{6} $ cm
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Answer:
Correct Answer: A
Solution:
- $ \angle DCB=15^{o} $ $ \angle CAB=45^{o} $ and $ \angle CDB=75^{o} $ Let $ BD=x $ and $ AD=35 $ cm. $ \tan \angle CAB=\frac{CB}{AB} $
Þ $ \tan 45^{o}=\frac{CB}{35+x} $ \ $ \tan 75^{o}=\frac{CB}{DB}=\frac{CB}{x} $
Þ $ CB=x\tan 75^{o} $ $ CB=(35+x)\tan 45^{o} $ = $ x\tan 75^{o} $
Þ $ x=\frac{35\tan 45^{o}}{\tan 75^{o}-\tan 45^{o}} $ = $ \frac{35}{\tan 75^{o}-1} $ But $ \cos 75^{o}=\frac{x}{CD} $ $ CD=\frac{x}{\cos 75^{o}} $ $ =\frac{1}{\cos 75^{o}}\times \frac{35}{\tan 75^{o}-1}=\frac{35}{\sin 75^{o}-\cos 75^{o}} $ $ =\frac{35}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}}=\frac{35}{\frac{2}{2\sqrt{2}}} $ $ =35\sqrt{2} $ cm.
BETA
