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

Trigonometric values for specific sine angles are given below,
$\begin{array}{cccccc}\theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }\\ Sin\theta & 0& \frac{1}{2}& \frac{1}{\sqrt{2}}& \frac{\sqrt{3}}{2}& 1\end{array}$

From $sinA=\frac{1}{2}$, we get $A={30}^{\circ }$----(1)

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

$sin(A+C)=sinB$ we get $A+C=B$ ------(3)

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

$\Rightarrow B+C={60}^{\circ },C={60}^{\circ }-{45}^{\circ }$
$\Rightarrow C={15}^{\circ }$

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