Chapter 8.6, Problem 49PPS

To determine

To calculate:

The perimeter of the given diagram.

Answer to Problem 49PPS

The perimeter of the quadrilateral is $275.1$ units.

Explanation of Solution

Given information:

  

Concept used:

Law of sines:

  $a=b\frac{\sin \alpha }{\sin \beta }$

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Calculation:

  

First name the quadrilateral $ABCD$ .

So, find the $BD$ using the law of cosines because as having the two sides and an included angle.

Law of cosines:

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos C\\ {\left(BD\right)}^2={\left(60\right)}^2+{\left(63\right)}^2-2\left(60\right)\left(63\right)\cos {48}^{\circ }\\ {\left(BD\right)}^2=3600+3696-5058.6\\ {\left(BD\right)}^2=2510.4\\ BD\approx 50.1\end{array}$

The sum of the angles of a triangle is $180$ . So,

  $m\angle ADB=180-\left(37+87\right)=56$

Now, use the law of sines to solve for $AD$ and $AB$ , as having the two angles and a nonincluded side.

Law of sines:

  $a=b\frac{\sin \alpha }{\sin \beta }$

First substitute,

  $\frac{\sin {37}^{\circ }}{50.1}=\frac{\sin {87}^{\circ }}{AD}$

Use the cross-product property,

  $AD\cdot \sin 37=55.1\cdot \sin 87$

Now, divide each side by $\sin 37$

  $AD=\frac{50.1\cdot \sin 87}{\sin 37}$

Now get the value of $AD$ ,

  $AD=83.1$

Similarly, solve for $AB$ , using the law of sines:

First substitute,

  $\frac{\sin {37}^{\circ }}{50.1}=\frac{\sin {56}^{\circ }}{AB}$

Use the cross products property,

  $AB\cdot \sin 37=50.1\cdot \sin 56$

Divide each side by $\sin 37$

  $AB=\frac{50.1\cdot \sin 56}{\sin 37}$

Now get the value of $AB$ ,

  $AB=69.0$

Therefore, the perimeter of the quadrilateral is about $63+60+83.1+69.0=275.1$ units.