Question

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

Find $sin(A+B-C)$.

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

From $sinA=\frac{1}{2}$, we get $A={30}^{\circ }----\left(1\right)$
$sin(B+C)=\frac{\sqrt{3}}{2}$, as we know that $B+C={60}^{\circ }-----\left(2\right)$
$sin(A+C)=sinB$
we get $A+C=B------\left(3\right)$
From the equations (1), (2) and (3)
$C={60}^{\circ }-B,A={30}^{\circ }$
$\Rightarrow A+C={60}^{\circ }-B+{30}^{\circ }=B$
$\Rightarrow 2\times B={90}^{\circ }$
$\Rightarrow B={45}^{\circ }$
$\Rightarrow B+C={60}^{\circ },C={60}^{\circ }-{45}^{\circ }$
$\Rightarrow C={15}^{\circ }$
$\Rightarrow A+B-C={30}^{\circ }+{45}^{\circ }-{15}^{\circ }={60}^{\circ }$
$\therefore sin(A+B-C)=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$