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Mathematics Question on Trigonometric Functions

If sin⁡(120−A)=sin⁡(120−B)\sin(120 - A)= \sin(120 - B) and 0<A,B<π0< A, B < \pi then all values of A,BA, B are given by.

A

A+B=π3A + B = \frac{\pi}{3}

B

A−BA - B or A+B=π3A+ B = \frac{\pi}{3}

C

A=BA = B

D

A+B=0A + B = 0

Answer

A−BA - B or A+B=π3A+ B = \frac{\pi}{3}

Explanation

Solution

Given trigonometrical equation is sin⁡(120−A)=sin⁡(120−B)\sin(120 -A)= \sin(120 - B)
Since sine is positive in II quadrant
∴\therefore either 120−A=120−B120 - A = 120 - B
⇒  A=B\Rightarrow \, \, A = B
or 120−A=180−(120−B) 120 - A = 180 -(120 -B)
⇒  120−A=60+B\Rightarrow \, \, 120 -A = 60 + B
⇒  A+B=π3\Rightarrow \, \, A+ B = \frac{\pi}{3}