Question
In △ABC, if $\mathrm{m} \angle \mathrm{A}=\frac{7 \pi^c}{36}, \mathrm{~m} \angle \mathrm{B}=120^{\circ}$, find $m \angle C$ in degree and radian.
✓
Answer
We know that $\theta^c=\left(\theta \times\left(\theta \times \frac{180}{\pi}\right)^{\circ}\right)^{\circ}$
$ \text { In } \triangle A B C_r$
$\mathrm{~m} \angle \mathrm{A}=\frac{7 \pi^c}{36}=\left(\frac{7 \pi}{36} \times \frac{180}{\pi}\right)^{\circ}=35^{\circ}$
$\mathrm{m} \angle B=120^{\circ}$
$\therefore \mathrm{m} \angle \mathrm{A}+\mathrm{m} \angle B+\mathrm{m} \angle \mathrm{C}=180^{\circ} $
$\ldots$ [Sum of the angles of a triangle is $180^{\circ}$ ]
$ \therefore 35^{\circ}+120^{\circ}+\mathrm{m} \angle \mathrm{C}=180^{\circ} \mathrm{m} \angle \mathrm{C}=180^{\circ}-35^{\circ}-120^{\circ}$
$\therefore \mathrm{m} \angle \mathrm{C}=25^{\circ} $
$ \text { In } \triangle A B C_r$
$\mathrm{~m} \angle \mathrm{A}=\frac{7 \pi^c}{36}=\left(\frac{7 \pi}{36} \times \frac{180}{\pi}\right)^{\circ}=35^{\circ}$
$\mathrm{m} \angle B=120^{\circ}$
$\therefore \mathrm{m} \angle \mathrm{A}+\mathrm{m} \angle B+\mathrm{m} \angle \mathrm{C}=180^{\circ} $
$\ldots$ [Sum of the angles of a triangle is $180^{\circ}$ ]
$ \therefore 35^{\circ}+120^{\circ}+\mathrm{m} \angle \mathrm{C}=180^{\circ} \mathrm{m} \angle \mathrm{C}=180^{\circ}-35^{\circ}-120^{\circ}$
$\therefore \mathrm{m} \angle \mathrm{C}=25^{\circ} $
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