Chapter 1.5, Problem 9PSA

To determine

To Find : Whether $\angle ABC$ has been trisected.

Explanation of Solution

Given information :

  $\begin{array}{l}\angle ABC=90°\\ \angle 1=\left(2x+10\right)°\\ \angle 2=\left(x+20\right)°\\ \angle 3=\left(3x\right)°\end{array}$

  

Proof:

Here, we have

  $\begin{array}{l}\angle ABC=90°\\ \angle 1=\left(2x+10\right)°\\ \angle 2=\left(x+20\right)°\\ \angle 3=\left(3x\right)°\\ \because \angle 1+\angle 2+\angle 3=\angle ABC=90°\\ \Rightarrow \left(2x+10\right)°+\left(x+20\right)°+\left(3x\right)°=90°\\ \Rightarrow \left(6x\right)°+30°=90°\\ \Rightarrow \left(6x\right)°=90°-30°\\ \Rightarrow \left(6x\right)°=60°\\ \Rightarrow x=10°\\ \therefore \angle 1=\left(2\times 10+10\right)°=30°\\ \angle 2=\left(10+20\right)°=30°\\ \angle 3=\left(3\times 10\right)°=30°\end{array}$

Since, the measure of each angle $\angle 1,\angle 2,\angle 3$ is equal.

  $\text{Hence, }\angle ABC\text{has been trisected}\text{.}$