Trigonometric Equations Question 223
Question: If $ \Delta =a^{2}-{{(b-c)}^{2}} $ , where $ \Delta $ is the area of triangle $ ABC $ , then tan A is equal to
[Pb. CET 1990; Kerala (Engg.) 2005]
Options:
A) $ \frac{15}{16} $
B) $ \frac{8}{15} $
C) $ \frac{8}{17} $
D) $ \frac{1}{2} $
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Answer:
Correct Answer: B
Solution:
- $ \Delta =2bc-(b^{2}+c^{2}-a^{2}) $ $ \Delta =2bc(1-\cos A)=2bc.2{{\sin }^{2}}\frac{A}{2} $ …..(i) $ \Delta =\frac{1}{2}bc\sin A=\frac{1}{2}(bc)2\sin \frac{A}{2}\cos \frac{A}{2} $ $ \Delta =bc\sin \frac{A}{2}\cos \frac{A}{2} $ …..(ii) \ $ \tan \frac{A}{2}=\frac{1}{4}=t $ , {by (i) and (ii)} $ \tan A=\frac{2t}{1-t^{2}}=\frac{8}{15} $ .
BETA
