Question

If $sin(A+B)=1,cos(A–B)=1$ and $0^{\circ }<A+B\le {90}^{\circ }$, find $A$ and $B.$

• A. ${60}^{\circ }$ , ${30}^{\circ }$
• B. ${45}^{\circ }$ , ${45}^{\circ }$
• C. ${30}^{\circ }$ , ${60}^{\circ }$
• D. ${15}^{\circ }$ , ${75}^{\circ }$

Answer 1

The correct option is B

${45}^{\circ }$ , ${45}^{\circ }$

Given that: $0^{\circ }<A+B\le {90}^{\circ }$

$sin(A+B)=1$ only when $A+B={90}^{\circ }$ ………… (1)

$cos(A–B)=1,$ only when $A-B=0^{\circ }$ ………… (2)

On adding (1) and (2) we get,
$2A={90}^{\circ }\Rightarrow A={45}^{\circ }$

Also, $A-B=0^{\circ }$
$B=A={45}^{\circ }$

Hence the value of both A and B is 45°.