Trigonometric Equations Question 224
Question: In $ \Delta ABC, $ $ a^{2}+b^{2}+c^{2}=ac+ab\sqrt{3}, $ then triangle is
[MP PET 2004]
Options:
A) Equilateral
B) Isosceles
C) Right angled
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- We have, $ a^{2}+b^{2}+c^{2}-ac-ab\sqrt{3}=0 $ $ \frac{a^{2}}{4}-,ac+c^{2}+\frac{3a^{2}}{4}+b^{2}-ab\sqrt{3}=0 $ $ {{[ \frac{a}{2}-c ]}^{2}}+{{[ \frac{\sqrt{3}a}{2}-b ]}^{2}}=0 $ i.e, $ a=2c $ and $ 2b=\sqrt{3}a $ i.e., $ b^{2}+c^{2}=a^{2} $ Hence triangle is right angled.
BETA
