Trigonometric Equations Question 288
Question: In a triangle $ ABC,b=\sqrt{3} $ , $ c=1 $ and $ \angle A=30^{o} $ , then the largest angle of the triangle is
[MP PET 2004]
Options:
A) $ 135^{o} $
B) $ 90^{o} $
C) $ 60^{o} $
D) $ 120^{o} $
Show Answer
Answer:
Correct Answer: D
Solution:
- We have, $ b=\sqrt{3},,c=1,,\angle A=30^{o} $ $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\Rightarrow \frac{\sqrt{3}}{2}=\frac{{{(\sqrt{3})}^{2}}+1^{2}-a^{2}}{2.\sqrt{3}.1} $
$ \therefore $ $ a=1,b=\sqrt{3},c=1 $ . b is the largest side. Therefore, the largest angle B is given by $ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{1+1-3}{2.1.1}=-\frac{1}{2} $ \ $ B=120{}^\circ $ .
BETA
