Trigonometric Equations Question 4
Question: In a triangle $ ABC, $ $ a=4,b=3 $ , $ \angle A=60^{o} $ . Then c is the root of the equation
[Roorkee 1993]
Options:
A) $ c^{2}-3c-7=0 $
B) $ c^{2}+3c+7=0 $
C) $ c^{2}-3c+7=0 $
D) $ c^{2}+3c-7=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \cos A=\frac{(b^{2}+c^{2}-a^{2})}{2bc} $
$ \Rightarrow \cos 60^{o}=\frac{1}{2}=\frac{9+c^{2}-16}{2.3c} $ Þ $ c^{2}-3c-7=0 $ .
BETA
