Question

If cos $(A-B)=sin(A+B)=\frac{1}{2}$,where A and B are positive, then smallest positive value of $A+B$ (in degrees) is

• A. ${45}^0$
• B. ${60}^0$
• C. ${105}^0$
• D. ${150}^0$

Answer 1

Answer: (D)

${150}^0$

$sin(A+B)=1/2$
therefore
$A+B=\frac{\pi }{6},\frac{5\pi }{6}$ ...(i)
$cos(A-B)=1/2$
Therefore
$A-B=\frac{\pi }{3},\frac{-\pi }{3}$ ...(ii)
Therefore
$A+B=\frac{\pi }{6}$ $A+B=\frac{5\pi }{6}$
$-(A-B)=\frac{\pi }{3}$ $-(A-B)=-\frac{\pi }{3}$
$B=-\frac{\pi }{6}$ $B=\frac{7\pi }{12}$ ...(iii)
Since both A and B are positive we proceed with (iii)
Therefore we get the Value of A by substituting in equation i
$A=\frac{\pi }{4}$
Therefore
$A+B=\frac{5\pi }{6}={150}^0$