Chapter 5, Problem 66RE

Solution

To calculate: The diagonals of a parallelogram with sides $15\text{ft}$ and $\text{24}\text{ft}$ .

The length of the diagonals of the parallelogram are $15.8\text{ft}$ and $36.8\text{ft}$ .

Given information: The length of the sides of the parallelogram are $15\text{ft}$ and $\text{24}\text{ft}$ . The given angle is $40°$ .

Formula used: For a triangle $ABC$, the law of cosines is given as follows:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Calculation:

Assume that the diagonals of the parallelogram be $x$ and $y$ .

It is known that the sum of two adjacent angles in the parallelogram is $180°$ . Find the other angle of the parallelogram.

  $180°-40°=140°$

Draw the parallelogram with given sides and angles as shown below.

  

Figure (1)

In $\Delta PQS$, apply the law of cosines and find the value of $y$ .

  $\begin{array}{c}{y}^2={\left(15\right)}^2+{\left(24\right)}^2-2\left(15\right)\left(24\right)\cos 40°\\ y^2=225+576-720\left(0.76\right)\\ y=\sqrt{801-720\left(0.76\right)}\\ y\approx 15.8\end{array}$

In $\Delta PRS$, apply the law of cosines and find the value of $x$ .

  $\begin{array}{c}{x}^2={\left(15\right)}^2+{\left(24\right)}^2-2\left(15\right)\left(24\right)\cos 140°\\ x^2=225+576-720\left(-0.76\right)\\ x=\sqrt{801-720\left(-0.76\right)}\\ x\approx 36.8\end{array}$

Thus, the length of the diagonals of the parallelogram are $15.8\text{ft}$ and $36.8\text{ft}$ .