Question

The perimeter of a $△ABC$ is 3 times the arithmetic mean of sine of its angles.

If the side ${}^{\prime }{b}^{\prime }=1$, then find the angle $B$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\pi }{2}$

$Given:$

$s=3\frac{sinA+sinB+sinC}{3}=sinA+sinB+sinC$

Multiplying $2R$ on both sides

$2Rs=2R(sinA+sinB+sinC).........................\left(1\right)$

we know that, $a=2RsinAb=2RsinBc=2RsinC$

Putting these values in $\left(1\right)$ we get,

$2R(a+b+c)=a+b+c\Rightarrow R=\frac{1}{2}$

Again from sine Rule,

$\frac{b}{sinB}=1\Rightarrow sinB=b$

Since, $b=1$(Given)

$sinB=1\Rightarrow B={90}^o$

Therefore, Answer is $\frac{\pi }{2}$