Trigonometric Equations Question 251
Question: If in a triangle $ ABC $ , $ b=\sqrt{3} $ , $ c=1 $ and $ B-C=90^{o} $ then $ \angle A $ is
[MP PET 1983]
Options:
A) $ 30^{o} $
B) $ 45^{o} $
C) $ 75^{o} $
D) $ 15^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \tan ( \frac{B-C}{2} )=\frac{b-c}{b+c}\cot \frac{A}{2} $
Þ $ \tan ( \frac{90{}^\circ }{2} )=\frac{\sqrt{3}-1}{\sqrt{3}+1}\cot \frac{A}{2} $
Þ $ \tan ( \frac{A}{2} )=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3} $
Þ $ \frac{A}{2}=15^{o}\Rightarrow A=30^{o} $ .
BETA
