Trigonometric Equations Question 313
Question: If in a triangle $ ABC $ , $ \cos A+\cos B+\cos C=\frac{3}{2} $ , then the triangle is
[IIT 1984]
Options:
A) Isosceles
B) Equilateral
C) Right angled
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Putting $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} $ in given relation, we get $ \frac{b^{2}+c^{2}-a^{2}}{2bc}+\frac{c^{2}+a^{2}-b^{2}}{2ac} $ $ +\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{3}{2} $
Þ $ a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2}) $ $ =a^{3}+b^{3}+c^{3}+3abc $
Þ $ {{(b-c)}^{2}}(b+c-a)+{{(c-a)}^{2}}(c+a-b) $ $ +{{(a-b)}^{2}}(a+b-c)=0 $ ?..(i) In triangle, $ b+c-a>0 $ etc. and hence (i) will hold good if each factor is zero so that $ a=b=c $ .
BETA
