Chapter 1.4, Problem 15PSC

To determine

To Show: $\angle ABC\cong \angle D$

Explanation of Solution

Given information :

  $\begin{array}{l}\angle 1=\left(x+7\right)°\\ \angle 2=\left(2x-3\right)°\\ \angle ABC=\left(x^2\right)°\\ \angle D=\left(5x-4\right)°\end{array}$

  

Proof:

Here, we have

  $\begin{array}{l}\angle 1=\left(x+7\right)°\\ \angle 2=\left(2x-3\right)°\\ \angle ABC=\left(x^2\right)°\\ \because \angle ABC=\angle 1+\angle 2\\ \Rightarrow \left(x^2\right)°=\left(x+7\right)°+\left(2x-3\right)°\\ \Rightarrow x^2-3x-4=0\\ \Rightarrow x^2-4x+x-4=0\\ \Rightarrow \left(x-4\right)\left(x+1\right)=0\\ \Rightarrow x=4,x=-1\end{array}$

  $\begin{array}{l}\text{For }x=4\\ \angle ABC=\left(4^2\right)°=16°\\ \angle D=\left(5\times 4-4\right)°=16°\\ \text{For }x=-1\\ \angle ABC=\left(-1^2\right)°=1\\ \angle D=\left(5\times \left(-1\right)-4\right)°=-9°\\ \therefore x=-1\text{is not the required solution}\text{.}\\ \text{Thus, for }x=4\\ \angle ABC\cong \angle D\end{array}$