Chapter 3.4, Problem 20WE

To determine

All three angles of triangle.

Answer to Problem 20WE

All three angles are, ${38}^0,{66}^0,{76}^0$

Explanation of Solution

Given :

One angle is 28 more than smallest angle and other is 2 times of smallest amgle.

Let the smallest angle is x.

Then other two angles will be $x+{28}^0$

And third angle will be, 2x

According to property of triangle sum of angles of a triangle is 180.

Therefore,

  $\begin{array}{l}x+x+{28}^0+2x={180}^0\\ 4x+{28}^0={180}^0\\ 4x={152}^0\\ x=\frac{{152}^0}{4}\\ ={38}^0\end{array}$

So the sides of triangle will be,

  $\begin{array}{l}x={38}^0\\ x+{28}^0={28}^0+{38}^0={66}^0\\ 2x=2\times {38}^0={76}^0\end{array}$

Therefore, angle of triangles are ${38}^0,{66}^0,{76}^0$