Question

In a triangle ABC, a = 3, b = 5, c = 7. Find the angle opposite to C.

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is C

${120}^{\circ }$

We know from trigonometric ratios of half angles that $tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{(s-a)\left(s\right)}}$
Here we need angle opposite to c i.e., C.
Semiperimeter, $S=\frac{a+b+c}{2}=\frac{3+5+7}{2}=\frac{15}{2}$
$tan\frac{C}{2}=\sqrt{\frac{(s-b)(s-a)}{(s-c)\left(s\right)}}$
$=\sqrt{\frac{(s-3)(s-5)}{(s-7)s}}=\sqrt{\frac{(15-6)(15-10)\left(2\right)}{4(15-14)\left(\frac{15}{2}\right)}}$
$=\sqrt{\frac{9\left(5\right)}{15}}=\sqrt{3}\Rightarrow \frac{C}{2}={60}^{\circ }\Rightarrow C={120}^{\circ }$