Question

The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.

Answer 1

Let the angles of a triangle be $x,yandz$

Since angles are in A.P and $x<y<z$

So, $2y=x+z-------\left(i\right)$

Step 1- Finding the value of the angle $y$

By Angle sum property of triangle

$$\begin{gathered} \angle x+\angle y+\angle z=180° \\ \Rightarrow y+2y=180\left(from\right(i\left)\right) \\ \Rightarrow 3y=180 \\ \Rightarrow y=\frac{180}{3} \\ \Rightarrow y=60° \end{gathered}$$

Step 2 - Finding the value of the angle $xandz$

Also given

$$\begin{gathered} \angle x+\angle y+\angle z=180° \\ \Rightarrow x+60+2x=180 \\ \Rightarrow 3x+60=180 \\ \Rightarrow 3x=180-60 \\ \Rightarrow 3x=120 \\ \Rightarrow x=\frac{120}{3} \\ \Rightarrow x=40° \end{gathered}$$

and

$$\begin{gathered} z=2x=2\times 40 \\ =80° \end{gathered}$$

Hence, the angles of a triangle are $40°,60°$° and $80°$