Trigonometric Equations Question 253
Question: The smallest angle of the triangle whose sides are $ 6+\sqrt{12},\sqrt{48},\sqrt{24} $ is
[EAMCET 1985]
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ A=6+\sqrt{12},b=\sqrt{48},c=\sqrt{24} $ . Clearly, c is the smallest side. Therefore, the smallest angle C is given by $ \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{\sqrt{3}}{2}\Rightarrow C=\frac{\pi }{6} $ .
BETA
