Question

Find $sin(A+B-C)$ if
$sin(B+C)=\frac{\sqrt{3}}{2}$
$sin(A+C)=sinB$ and
$sinA=\frac{1}{2}.$

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

Answer 1

The correct option is B

$\frac{\sqrt{3}}{2}$

Given,
$sin(B+C)=\frac{\sqrt{3}}{2}sin(A+C)=sinBsinA=\frac{1}{2}$

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

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

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

From the equations (1),(2) and (3)
$C={60}^{\circ }-B,A={30}^{\circ }$
$\Rightarrow A+C={30}^{\circ }+{60}^{\circ }-B=B$
$\Rightarrow 2\times B={90}^{\circ }$
$\Rightarrow B={45}^{\circ }$
$\Rightarrow B+C={60}^{\circ },C={60}^{\circ }-{45}^{\circ }$
$\Rightarrow C={15}^{\circ }$
$\Rightarrow A+B-C={30}^{\circ }+{45}^{\circ }-{15}^{\circ }={60}^{\circ }$
$\therefore sin(A+B-C)=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$