Chapter 8.6, Problem 47PPS

To determine

To calculate:

The perimeter of the given diagram.

Answer to Problem 47PPS

The perimeter of the triangle is about $24.3$ units.

Explanation of Solution

Given information:

  

Concept used:

Law of sines:

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

Calculation:

  

The sum of angle of a triangle that is $180$ . So, measure of the third angle is

  $180-\left(31+86\right)=63$

Use the law of sines to get the value of the missing sides, as having the two angles and nonincluded side.

Law of sines:

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

First substitute,

  $\frac{\sin {63}^{\circ }}{9}=\frac{\sin {86}^{\circ }}{x}$

Use the cross-product property,

  $x.\sin 63=9.\sin 86$

Now, divide each side by $\sin 63$

  $x=\frac{9.\sin 86}{\sin 63}$

Now get the value of $x$ ,

  $x=10.1$

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

First substitute,

  $\frac{\sin {63}^{\circ }}{9}=\frac{\sin {31}^{\circ }}{y}$

Use the cross products property,

  $y.\sin 63=9.\sin 31$

Divide each side by $\sin 63$

  $y=\frac{9\cdot \sin 31}{\sin 63}$

Now get the value of $y$ ,

  $y=5.2$

Therefore, the perimeter of the triangle is about $9+10.1+5.2=24.3$ units.