Question

In $△ABC,\angle A+\angle B={70}^{\circ }$ and $\angle B+\angle C={135}^{\circ }$. Find the measure of each angle of the triangle.

• A. ${45}^{\circ },{25}^{\circ },{110}^{\circ }$
• B. ${25}^{\circ },{25}^{\circ },{110}^{\circ }$
• C. ${45}^{\circ },{35}^{\circ },{110}^{\circ }$
• D. None of these

Answer 1

The correct option is A ${45}^{\circ },{25}^{\circ },{110}^{\circ }$

Sum of angle of triangle $△ABC$ is ${180}^o$.

So, $\angle A+\angle B+\angle C={180}^o$

Now,

$\angle A+\angle B={70}^o$..........$\left(1\right)$

$\angle B+\angle C={135}^o$..........$\left(2\right)$

Adding equations $(1+2)$:

$\angle A+\angle B+\angle C+\angle B={135}^o+{70}^o$

$=>{180}^o+\angle B={205}^o$...................[$\angle A+\angle B+\angle C={180}^o$]

$=>\angle B={25}^o$

Now;

$\angle A+\angle B={70}^o$..........$\left(1\right)$

$=>\angle A={70}^o-{25}^o={45}^o$

And

$\angle B+\angle C={135}^o$..........$\left(2\right)$

$=>\angle C={135}^o-{25}^o={110}^o$

$\therefore \angle A={45}^o,\angle B={12}^o,\angle C={110}^o$