Question

In $△ABC,\angle A=\frac{2{\pi }^c}{3}$ and $\angle B={45}^o$. Find $\angle C$ in both the systems.

Answer 1

Given

$\angle A={\frac{2\pi }{3}}^c$

$\angle B={45}^0$

Coverting these into different systems we get,

$\Rightarrow \angle A={\frac{2\pi }{3}}^c=\frac{2\times {180}^0}{3}$

$\Rightarrow \angle A={120}^0$

$\Rightarrow \angle B={45}^0={45}^0\times \frac{{\pi }^c}{{180}^0}$

$\Rightarrow \angle B=\frac{{\pi }^c}{4}$

We know that in a triangle,

$\angle A+\angle B+\angle C={180}^0$

$\Rightarrow {120}^0+{45}^0+\angle C={180}^0$

$\Rightarrow {165}^0+\angle C={180}^0$

$\Rightarrow \angle C={180}^0-{165}^0$

$\Rightarrow \angle C={15}^0$

$\Rightarrow \angle C={15}^0\times \frac{{\pi }^c}{{180}^0}$

$\Rightarrow \angle C=\frac{{\pi }^c}{12}$

Therefore, $\Rightarrow \angle C={15}^0=\frac{{\pi }^c}{12}$