Question

One of the exterior angles of a triangle is ${80}^{\circ }$ and the interior opposite angles are in the ratio 3 : 5. Find the angles of the triangle.

Answer 1

Let $\angle ACX$ be the exterior angle of $\Delta ABC$ at C such that $\angle ACX={80}^{\circ }$. Clearly, $\angle Aand\angle B$ are the interior opposite angles.
It is given that $\angle A:\angle B=3:5.$ So, let $\angle A=3x^{\circ }and\angle B=5x^{\circ }$.
$\angle ACX=\angle A+\angle B$
$\Rightarrow {80}^{\circ }=3x+5x$ [By exterior angle property]
$\Rightarrow 8x={80}^{\circ }$ [Dividing both sides by 8]
$\Rightarrow x={10}^{\circ }$
$\angle A=3x^{\circ }={30}^{\circ }and\angle B=5x^{\circ }={50}^{\circ }$


Now, $\angle A+\angle B+\angle C={180}^{\circ }$ (Angle sum property of triangle)
$\Rightarrow {30}^{\circ }+{50}^{\circ }+\angle C={180}^{\circ }$
$\Rightarrow {80}^{\circ }+\angle C={180}^{\circ }$
$\Rightarrow \angle C={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
Hence, $\angle A={30}^{\circ },\angle B={50}^{\circ }and\angle C={100}^{\circ }$