Chapter 11, Problem 2STP

To determine

To find:the angle $\angle MLN$ of the given $\Delta DEF$ .

Answer to Problem 2STP

  $\angle MLN={42}^{\circ }$ .

Explanation of Solution

Given:

The right triangle $\Delta DEF$ is congruent to the right triangle is $\Delta LMN$ .

  $\angle EDF={42}^{\circ }$

Calculation:

According to the question,

  

As the right triangle $\Delta DEF$ is congruent to the right triangle is $\Delta LMN$ .

  $\Rightarrow \Delta DEF\cong \Delta LMN$ .

Therefore, by corresponding parts of congruent triangles (CPCT),

  $\begin{array}{l}\Rightarrow \angle DEF=\angle LMN\\ \angle EDF=\angle MLN\\ \angle EDF={42}^{\circ }\left\{given\right\}\\ \angle MLN={42}^{\circ }\end{array}$

Hence, theangle $\angle MLN={42}^{\circ }$ .