Find out the angles in a given triangle

Question

In a $\Delta ABC$, $a=7$, $c=9$ & $\angle A=36^\circ$. The values of $\angle B$ & $\angle C$ are

a.) $94.91^\circ$ & $49.09^\circ$

b.) $95.4^\circ$ & $48.6^\circ$

c.) $13.09^\circ$ & $130.91^\circ$

d.) both a & c

I have applied Sine Rule as follows $$\frac{\sin A}{a}=\frac{\sin C}{c} $$$$\sin C=\frac{9}{7}\sin 36^\circ \implies C\approx 49.09^\circ$$ Thus I obtained $\angle B=94.91^\circ$ & $\angle C=49.09^\circ$ & found option (a) is correct, but my book gives option d.) as correct. Could any please explain?

I am 12th grade. Thanks for your help in advance!

Answer 1

This figure below explains everything.

Answer 2

If you draw a picture for the given data, you'll see you've got a "side-side-angle" situation, which in general does not have a unique solution. That's why in this case there are two possibilities instead of just one.

(Minor TeX remark: "^\circ" does a better job than "^o" of producing the degree symbol.)

Answer 3

Remember that $\sin(a) = \sin(180-a)$.

$\sin C=\frac{9}{7}\sin 36^o \implies C\approx 49.09^o $

Therefore if $\sin(C) = \frac{9}{7}\sin 36^o$, then $C$ can be either $49.09^o$ or $180^o-49.09^o =130.91^o $.