Question

An exterior angle of a triangle measures $110°$ and its interior opposite angles are in the ratio $2:3$. Find the angles of the triangle.

Answer 1

Exterior Angle Theorem.

When one of the sides of a triangle is extended, the external angle formed between the extended side and its adjacent side is called an exterior angle.

The Exterior Angle Theorem states that “The exterior angle is equal to the sum of interior opposite angles of a triangle”.

Consider the $∆\mathrm{ABC}$. Draw an exterior angle to the $∆\mathrm{ABC}$ by extending the side $\mathrm{BC}$.

Given the exterior angle of a triangle is $110°$ and its interior opposite angles are in the ratio $2:3$.

Therefore, $\angle \mathrm{ACD}=110°$ and $\angle \mathrm{A}:\angle \mathrm{B}=2:3$.

Let $\angle \mathrm{A}=2\alpha$ and $\angle \mathrm{B}=3\alpha$.

Hence, by the Exterior Angle Theorem,

$\begin{array}{rcl}\angle \mathrm{ACD}& =& \angle \mathrm{A}+\angle \mathrm{B}\\ & \Rightarrow & 110°=2\alpha +3\alpha \\ & \Rightarrow & 110°=5\alpha \\ & \Rightarrow & \alpha =\frac{110°}{5}\\ & \Rightarrow & \alpha =22°\end{array}$

Therefore,

$\begin{array}{rcl}\angle \mathrm{A}& =& 2\alpha \\ & \Rightarrow & \angle \mathrm{A}=2\times 22°\\ & \Rightarrow & \angle \mathrm{A}=44°\end{array}$

and

$\begin{array}{rcl}\angle \mathrm{B}& =& 3\alpha \\ & \Rightarrow & \angle \mathrm{B}=3\times 22°\\ & \Rightarrow & \angle \mathrm{B}=66°\end{array}$

Since the sum of all the angles of a triangle is equal to $180°$, therefore
$\begin{array}{rcl}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}& =& 180°\\ & \Rightarrow & 44°+66°+\angle \mathrm{C}=180°\\ & \Rightarrow & 110°+\angle \mathrm{C}=180°\\ & \Rightarrow & \angle \mathrm{C}=180°-110°\\ & \Rightarrow & \angle \mathrm{C}=70°\end{array}$

Therefore, the angles of the triangle are $44°$, $66°$ and $70°$.