Question

An exterior angle of a triangle is ${100}^{\circ }$ and its interior opposite angles are equal to each other. Find the measure of each angle of the triangle.

Answer 1

Hint:
We will take the unknown measures as $x^{\circ }$. Then we will equate the sum of interior angles to the exterior angle and find the unknown interior angles. Finally, we will use the angle sum property of a triangle to find the third angle.

Complete step by step solution:
Let us consider $△ABC$. Let us assume that the exterior angle is formed by producing the side $BC$ along the line $CD$.
We are given that the exterior angle is ${100}^{\circ }$. Let us take $\mathrm{\angle }ACD={100}^{\circ }$.
Also, it is given that the interior opposite angles are equal. Let us take each of them to be $x^{\circ }$i.e., $\mathrm{\angle }BAC=\mathrm{\angle }ABC=x^{\circ }$.

We know that “If the side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles”.
Applying this to $△ABC$, we get
$\mathrm{\angle }BAC+\mathrm{\angle }ABC={100}^{\circ }$……………………..$\left(1\right)$
Substituting $\mathrm{\angle }BAC=\mathrm{\angle }ABC=x^{\circ }$ in equation $\left(1\right)$, we get
$\Rightarrow x^{\circ }+x^{\circ }={100}^{\circ }$
Adding the like terms on the LHS, we get
$\Rightarrow 2x^{\circ }={100}^{\circ }$
Dividing both sides by 2, we get
$\Rightarrow x^{\circ }={\displaystyle \frac{{100}^{\circ }}{2}}={50}^{\circ }$
Therefore, $\mathrm{\angle }BAC=\mathrm{\angle }ABC={50}^{\circ }$. Now, we will use the angle sum property of a triangle to find the third angle. We know that “Sum of the angles of a triangle is ${180}^{\circ }$”.
Applying this to $△ABC$,
$\mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }ACB={180}^{\circ }$ ……………………………………$\left(2\right)$
Substituting $\mathrm{\angle }BAC=\mathrm{\angle }ABC={50}^{\circ }$ in equation $\left(2\right)$, we get,
$\Rightarrow {50}^{\circ }+{50}^{\circ }+\mathrm{\angle }ACB={180}^{\circ }$
Adding the like terms, we get
$\Rightarrow {100}^{\circ }+\mathrm{\angle }ACB={180}^{\circ }$
Subtracting ${100}^{\circ }$ on both the sides, we get
$\Rightarrow \mathrm{\angle }ACB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Thus, the angles of $△ABC$ are ${50}^{\circ },{50}^{\circ }\text{ and }{100}^{\circ }$.

Note:
We can also find $\mathrm{\angle }ACB$ as follows:
Since $BD$ is a line, $\mathrm{\angle }ACB\text{ and }\mathrm{\angle }ACD$ form a linear pair.
Now, we know that the sum of a linear pair of angles is ${180}^{\circ }$. Therefore,
$\mathrm{\angle }ACB+\mathrm{\angle }ACD={180}^{\circ }$
Substituting $\mathrm{\angle }ACD={100}^{\circ }$ in the above equation, we get
$\mathrm{\angle }ACB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
Now again using angle sum property, we get
$\mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }ACB={180}^{\circ }$
Substituting the values in the above equation, we get
$\Rightarrow x+x+{80}^{\circ }={180}^{\circ }$
Adding and subtracting the like terms, we get
$$\begin{gathered} \Rightarrow 2x={180}^{\circ }-{80}^{\circ } \\ \Rightarrow 2x={100}^{\circ } \end{gathered}$$
Dividing both side by 2, we get
$\Rightarrow x={50}^{\circ }$
Thus, the angles of $△ABC$ are ${50}^{\circ },{50}^{\circ }\text{ and }{100}^{\circ }$.