Trigonometric Equations Question 320
Question: In a triangle $ ABC $ , $ AD $ is altitude from A. Given $ b>c, $ $ \angle C=23^{o} $ and $ AD=\frac{abc}{b^{2}-c^{2}}, $ then $ \angle B= $
[IIT 1994]
Options:
A) $ 67^{o} $
B) $ 44^{o} $
C) $ 113^{o} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{a^{2}-(b^{2}-c^{2})}{2ac} $ Now, $ AD=\frac{abc}{b^{2}-c^{2}} $ ; \ $ \cos B=\frac{a^{2}-\frac{abc}{AD}}{2ac} $ Also, $ AD=b\sin 23^{o} $ ; \ $ \cos B=\frac{a-\frac{c}{\sin 23^{o}}}{2c} $ By sine formula, $ \frac{a}{c}=\frac{\sin (B+23^{o})}{\sin 23^{o}} $ \ $ \cos B=( \frac{\sin (B+23^{o})}{\sin 23^{o}}-\frac{1}{\sin 23^{o}} )\div 2 $
Þ $ \sin (23^{o}-B)=-1=\sin (-90^{o}) $ \ $ 23^{o}-B=-90^{o} $ or $ B=113^{o} $ .
BETA
