Chapter 14.4, Problem 33PPSE

Solution

a.

To calculate: The value of $m\angle D$ , $m\angle ABD$ and $m\angle ABC$ if $m\angle A=30°$ , $AB=10$ and $BC=BD=6$ using the law of sines.

  

The value of $m\angle D$ is $56.4°$ , $m\angle ABD$ is $93.6°$ and $m\angle ABC$ is $26.4°$ .

Given information:

The given values of the triangle are $m\angle A=30°$ , $AB=10$ and $BC=BD=6$ .

  

Formula Used:

Use the law of sines.

Calculation :

To calculate $m\angle D$ , use $\Delta ABD$ and law of sines.

  $\begin{array}{c}\frac{\sin D}{10}=\frac{\sin 30°}{6}\\ \sin D=\frac{10\sin 30°}{6}\\ m\angle D={\sin }^{-1}\left(\frac{10\sin 30°}{6}\right)\\ m\angle D\approx 56.4°\end{array}$

The sum of interior angles of $\Delta ABD$ is as follows.

  $\begin{array}{c}\angle A+\angle ABD+\angle D=180°\\ 30+\angle ABD+56.4=180°\\ m\angle ABD\approx 93.6°\end{array}$

As $\Delta BCD$ is an isosceles triangle, then $\angle BCD=\angle D=56.4°$ , which follows that $\angle CBD=67.2°$ .

Using $\angle ABD$ , write

  $\begin{array}{c}\angle ABD=\angle ABC+\angle CBD\\ 93.6°=\angle ABC+67.2°\\ m\angle ABC\approx 26.4°\end{array}$

Thus, the value of $m\angle D$ is $56.4°$ , $m\angle ABD$ is $93.6°$ and $m\angle ABC$ is $26.4°$ .

b.

To determine: If $\Delta EFG$ be congruent to $\Delta ABD$ if $EF=10$ , $FG=6$ and $\angle E\cong \angle A$ if two sides and a nonincluded angle of $\Delta ABC$ are congruent to the corresponding part of $\Delta ABD$ .

The triangle $\Delta EFG$ is not congruent to $\Delta ABD$ .

Given information:

The two sides and a nonincluded angle of $\Delta ABC$ are congruent to the corresponding part of $\Delta ABD$ .

Explanation :

The triangle $\Delta EFG$ might be congruent to $\Delta ABC$ as they are also congruent.

The reason is similar to the statement that says about $\Delta ABC$ and $\Delta ABD$ not being congruent