Question

The sum of all exterior angles of a triangle is
A.$${360}^0$$
B.$${180}^0$$
C.$${540}^0$$
D.None of these

Answer 1

Hint: Assume a $$\mathrm{\Delta }ABC$$ in which $$\mathrm{\angle }A=\alpha$$ , $$\mathrm{\angle }B=\beta$$ , and $$\mathrm{\angle }C=\gamma$$ . We know the property that the measure of an exterior angle of a triangle is equal to the sum of the opposite interior angles. Now, using the property we can say that the exterior angle $$\mathrm{\angle }ACP$$ is equal to the summation of the angles $$\alpha$$ and $$\beta$$ , $$\mathrm{\angle }BAQ$$ is equal to the summation of the angles $$\gamma$$ and $$\beta$$ , and $$\mathrm{\angle }CBR$$ is equal to the summation of the angles $$\alpha$$ and $$\gamma$$ .

Complete step-by-step answer:

Assume a $$\mathrm{\Delta }ABC$$ in which $$\mathrm{\angle }A=\alpha$$ , $$\mathrm{\angle }B=\beta$$ , and $$\mathrm{\angle }C=\gamma$$ .
For the angles $$\alpha$$ and $$\beta$$ , $$\mathrm{\angle }ACP$$ is an exterior angle.
We know the property that the measure of an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Now, using the property we can say that the exterior angle $$\mathrm{\angle }ACP$$ is equal to the summation of the angles $$\alpha$$ and $$\beta$$ .
$$\mathrm{\angle }ACP=\alpha +\beta$$ …………………(1)
For the angles $$\gamma$$ and $$\beta$$ , $$\mathrm{\angle }BAQ$$ is an exterior angle.
We know the property that the measure of an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Now, using the property we can say that the exterior angle $$\mathrm{\angle }BAQ$$ is equal to the summation of the angles $$\gamma$$ and $$\beta$$ .
$$\mathrm{\angle }BAQ=\gamma +\beta$$ …………………(2)
For the angles $$\alpha$$ and $$\gamma$$ , $$\mathrm{\angle }CBR$$ is an exterior angle.
We know the property that the measure of an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Now, using the property we can say that the exterior angle $$\mathrm{\angle }CBR$$ is equal to the summation of the angles $$\alpha$$ and $$\gamma$$ .
$$\mathrm{\angle }CBR=\alpha +\gamma$$ …………………(3)
For the $$\mathrm{\Delta }ABC$$ exterior angles are $$\mathrm{\angle }CBR$$ , $$\mathrm{\angle }BAQ$$ , and $$\mathrm{\angle }ACP$$ .
Now, the sum of all exterior angles,
$$\mathrm{\angle }ACP+\mathrm{\angle }BAQ+\mathrm{\angle }CBR$$ ………………………(4)
From equation (1), equation (2), equation (3), and equation (4), we get
$$\begin{array}{rl}& \mathrm{\angle }ACP+\mathrm{\angle }BAQ+\mathrm{\angle }CBR\\ & =\alpha +\beta +\gamma +\beta +\alpha +\gamma \end{array}$$
$$=2(\alpha +\beta +\gamma )$$ ………………(5)
We know that the sum of all interior angles of a triangle is $${180}^0$$ .
That is,
$$(\alpha +\beta +\gamma )={180}^0$$ …………………..(6)
From equation (5) and equation (6), we get
$$\begin{array}{rl}& =2(\alpha +\beta +\gamma )\\ & =2\times {180}^0\\ & ={360}^0\end{array}$$
So, the sum of all exterior angles of a triangle is $${360}^0$$ .
Hence, the correct option is A.

Note: We can also solve this question by using linear pair angles.

Assume a $$\mathrm{\Delta }ABC$$ in which $$\mathrm{\angle }A=\alpha$$ , $$\mathrm{\angle }B=\beta$$ , and $$\mathrm{\angle }C=\gamma$$ .
Here, $$\mathrm{\angle }ACB$$ and $$\mathrm{\angle }ACP$$ are linear pair angles.
So, $$\mathrm{\angle }ACB+\mathrm{\angle }ACP={180}^0$$
$$\mathrm{\angle }ACP={180}^0-\gamma$$ ……………(1)
Here, $$\mathrm{\angle }BAQ$$ and $$\mathrm{\angle }BAC$$ are also linear pair angles.
So, $$\mathrm{\angle }BAQ+\mathrm{\angle }BAC={180}^0$$
$$\mathrm{\angle }BAQ={180}^0-\alpha$$ ……………(2)
Here, $$\mathrm{\angle }CBR$$and $$\mathrm{\angle }CBA$$ are linear pair angles.
So, $$\mathrm{\angle }CBR+\mathrm{\angle }CBA={180}^0$$
$$\mathrm{\angle }CBR={180}^0-\beta$$ ……………(3)
From equation (1), equation (2), and equation (3),
$$\begin{array}{rl}& \mathrm{\angle }ACP+\mathrm{\angle }BAQ+\mathrm{\angle }CBR\\ & ={180}^0-\gamma +{180}^0-\alpha +{180}^0-\beta \end{array}$$
$$={540}^0-(\alpha +\beta +\gamma )$$ ………………….(4)
We know that the sum of all interior angles of a triangle is $${180}^0$$ .
That is,
$$(\alpha +\beta +\gamma )={180}^0$$ ………………….(5)
From equation (4) and equation (5),
$$\begin{array}{rl}& ={540}^0-(\alpha +\beta +\gamma )\\ & ={540}^0-{180}^0\\ & ={360}^0\end{array}$$
So, the sum of all exterior angles of a triangle is $${360}^0$$ .