Question

If $sin(B+C)=\frac{\sqrt{3}}{2}$, $sin(A+C)=sinB\text{and}$ $sinA=\frac{1}{2}.$, then $\sin (A+B-C)$ is .

• A. $1$
• B. $\sqrt{\frac{1}{4}}$
• C. $\sqrt{\frac{3}{4}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\sqrt{\frac{3}{4}}$

Given,
$sinA=\frac{1}{2}$,
$\Rightarrow A={30}^{\circ }...\left(1\right)$

$sin(B+C)=\frac{\sqrt{3}}{2}$
$\Rightarrow B+C={60}^{\circ }...\left(2\right)$

$sin(A+C)=sinB$
$\Rightarrow A+C=B...\left(3\right)$

From the equations (1),(2) and (3)
$C={60}^{\circ }-B,A={30}^{\circ }$
$\Rightarrow A+C={60}^{\circ }-B+{30}^{\circ }=B$
$\Rightarrow 2\times B={90}^{\circ }$
$\Rightarrow B={45}^{\circ }$

$\text{Also,}B+C={60}^{\circ }$
$\Rightarrow C={60}^{\circ }-{45}^{\circ }$
$\Rightarrow C={15}^{\circ }$
$\Rightarrow A+B-C={30}^{\circ }+{45}^{\circ }-{15}^{\circ }={60}^{\circ }$
$\therefore sin(A+B-C)=sin{60}^{\circ }=\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{4}}$