Chapter 8.6, Problem 32PPS

To determine

To calculate:

The triangle and angle measures degree and side measures to the nearest tenth.

Answer to Problem 32PPS

The measurements of triangle are $m\angle E\approx {41}^{\circ },m\angle F\approx {31}^{\circ },DF\approx 10.1$ .

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:

  

With the help of the above diagram it is easy to get the two sides and an nonincluded angle. So, with the help of the Law of sines solve the issue.

Law of sines:

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

First substitute,

  $\frac{\sin F}{7.9}=\frac{\sin 108}{14.6}$

Cross products property,

  $14.6\cdot \left(\sin F\right)=7.9\cdot \left(\sin 108\right)$

Divide each side by $14.6$

  $\sin F=\frac{7.9\cdot \sin 108}{14.6}$

Now, get the value of $F$ ,

  $F\approx 31$

Now, calculate the sum of angles of triangles is $180$ . So,

  $m\angle E=180-\left(108+31\right)=41$

Since, as know the two sides and an included angle, use the law of cosines to get $DF$ .

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

So,

  $\begin{array}{l}{\left(DF\right)}^2={\left(7.9\right)}^2+{\left(14.6\right)}^2-2\left(7.9\right)\left(14.6\right)\cos \left(41\right)\\ {\left(DF\right)}^2=101.47\\ DF=\sqrt{101.47}\\ DF=10.1\end{array}$

Therefore, the measurements of triangle are $m\angle E\approx {41}^{\circ },m\angle F\approx {31}^{\circ },DF\approx 10.1$