Question

If $\sin (A+B+C)=1$, $\sec (A+C)=2$ and $A,B,C$ being acute angles, then $B=$

• A. ${90}^0$
• B. ${45}^0$
• C. ${30}^0$
• D. ${60}^0$

Answer 1

Answer: (C)

${30}^0$

Given $\sin (A+B+C)=1$ and $\sec (A+C)=2,\cos (A+C)=\frac{1}{2}$

$A,B,C$ acute angles.

$A+B+C=(2n+1)\frac{\pi }{2}$
$\Rightarrow A+C={60}^o$

Put $n=0$
$A+B+C=\frac{\pi }{2}$

$B=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}={30}^o$