Chapter 6.6, Problem 24E

To determine

To find: The measure of angles of parallelogram.

Answer to Problem 24E

The measure of angles of parallelogramis $B=97.9,D=97.9,A=104.8,C=104.8$ .

Explanation of Solution

Given information: The side of parallelogram is 20 and 32 centimeters.

Calculation:

Consider the figure.

  

Consider the triangle ABC.

  $\begin{array}{c}{b}^2=c^2+a^2-2ca\cos B\\ {\left(40\right)}^2={\left(20\right)}^2+{\left(32\right)}^2-2\left(20\right)\left(32\right)\cos B\\ B={\cos }^{-1}\left(-0.1375\right)\\ =97.9\end{array}$

Angle B and D are equal by property of parallelogram.

Consider the triangle ABC.

  $\begin{array}{c}{a}^2=b^2+c^2-2cb\cos A\\ {\left(32\right)}^2={\left(40\right)}^2+{\left(20\right)}^2-2\left(40\right)\left(20\right)\cos A\\ A={\cos }^{-1}\left(0.61\right)\\ =54.4\end{array}$

As the diagonal bisects the angle, the angle is $2\times 52.4=104.8$

Angle A and C are equal by property of parallelogram.

Thus, the measure of angles of parallelogram is $B=97.9,D=97.9,A=104.8,C=104.8$ .