Trigonometric Equations Question 117
Question: In $ \Delta ABC, $ if $ 8R^{2}=a^{2}+b^{2}+c^{2}, $ then the triangle is
Options:
A) Right angled
B) Equilateral
C) Acute angled
D) Obtuse angled
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 8R^{2}=a^{2}+b^{2}+c^{2}=4R^{2}({{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C) $
$ \Rightarrow $ $ {{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2 $
$ \Rightarrow $ $ ({{\cos }^{2}}A-{{\sin }^{2}}C)+{{\cos }^{2}}B=0 $
$ \Rightarrow $ $ \cos (A-C)\cos (A+C)+{{\cos }^{2}}B=0 $
Þ $ 2\cos A\cos B\cos C=0 $ So that, $ \cos A=0 $ or $ \cos B=0 $ or $ \cos C=0 $
$ \Rightarrow $ $ A=\frac{\pi }{2} $ or $ B=\frac{\pi }{2} $ or $ C=\frac{\pi }{2} $ .
BETA
