Question

In $\Delta ABC,if\cos \frac{A}{2}=\sqrt{\frac{b+c}{2c}},$ then

• A. area of triangle is $\frac{1}{2}ab$
• B. circumradius is equal to $\frac{1}{2}c$
• C. area of triangle is $\frac{1}{2}bc$
• D. circumradius is equal to $\frac{1}{2}a$

Answer 1

Answer: (A), (B)

area of triangle is $\frac{1}{2}ab$
circumradius is equal to $\frac{1}{2}c$

Given, $\cos \frac{A}{2}=\sqrt{\frac{b+c}{2c}}$

$\Rightarrow {\cos }^2\frac{A}{2}=\frac{b+c}{2c}=1+\cos A$

$\Rightarrow \cos A=\frac{b}{c}$

$\therefore \angle C={90}^0$

Hence area of triangle $ABC$ is $=\frac{1}{2}ab$
and circumradius, $R=\frac{c}{2\sin C}=\frac{c}{2}$