MHB Using trigonometry to prove <c=50 degree

[Albert1]

$\triangle ABC,\angle B=30^o$ and
$\overline{BC}^2-\overline{AB}^2=\overline{AB}\times \overline{AC}$
using trigonometry to prove $\angle C=50^o$

 

[kaliprasad]

$\triangle ABC,\angle B=30^o$ and
$\overline{BC}^2-\overline{AB}^2=\overline{AB}\times \overline{AC}$
using trigonometry to prove $\angle C=50^o$

Spoiler

we have $\angle B=30^\circ$ hence $\angle A + \angle C=150\circ$ so we caan choose
$\angle A=(75^\circ+\theta)$ and $\angle C=(75^\circ-\theta)$
from the given equation using lay of sines we get
$\sin ^2\angle A - \ sin ^2 \angle C = \sin \angle C \sin\angle B$
or $( \sin \angle A + \sin \angle C)(sin \angle A - \sin \angle C) = \frac{1}{2}\sin \angle C$
or $(\sin ((75^\circ+ \theta) + \sin ((75^\circ - \theta)) (\sin ((75^\circ+ \theta) - \sin ((75^\circ - \theta)) = \frac{1}{2}\sin \angle C$
or $(2 \sin\, 75^\circ \cos \theta)(2 \cos 75^\circ\sin \theta) = \frac{1}{2}\sin \angle C$
or $sin \, 150^\circ \sin 2\theta = \frac{1}{2}\sin \angle C$
or $\frac{1}{2} \sin 2\theta = \frac{1}{2}\sin \angle C$
so $2 \theta = C$ or $2\theta = 180^\circ - C$
from $2\theta = 75 - \theta$ we get $\theta = 25^\circ$ or $C= 50^\circ$
from $2\theta = 180 - (75 - \theta) $ we get $\theta = 105^\circ$ out side limit as it has to be less than $75^\circ$
so
$C= 50^\circ$
proved