Question

If $\sin(120 - A)= \sin(120 - B)$ and $0< A, B < \pi $ then all values of $A, B$ are given by.

• A. $A + B = \frac{\pi}{3}$
• B. $A - B$ or $A+ B = \frac{\pi}{3}$
• C. $A = B$
• D. $ A + B = 0 $

Answer 1

Answer: (B)

Given trigonometrical equation is $\sin(120 -A)= \sin(120 - B)$
Since sine is positive in II quadrant
$\therefore $ either $120 - A = 120 - B$
$\Rightarrow \, \, A = B$
or $ 120 - A = 180 -(120 -B)$
$ \Rightarrow \, \, 120 -A = 60 + B$
$ \Rightarrow \, \, A+ B = \frac{\pi}{3}$