Chapter 8.6, Problem 57HP

(a)

To determine

To draw:

The triangle using the Law of Sines

Answer to Problem 57HP

The triangle using law of sine:

  

Explanation of Solution

Given information:

Law of Sines:

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Interpretation:

If $\Delta ABC$ has length $a$ , $b$ and $c$ representing the length of sides opposite angles with measure $A$ , $B$ and $C$ :

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\mathrm{...}\left(1\right)$

Put the value in equation (1):

  $\frac{\sin {31}^{\circ }}{7.9}=\frac{\sin {108}^{\circ }}{EF}$

The property of cross products:

  $\begin{array}{l}\left(EF\right)\sin {31}^{\circ }=7.9\left(\sin {108}^{\circ }\right)\\ EF=\frac{7.9\sin {108}^{\circ }}{\sin {31}^{\circ }}\\ EF\approx 14.6\end{array}$

Also, the law of sine is:

  $\frac{\sin {31}^{\circ }}{7.9}=\frac{\sin {41}^{\circ }}{DF}$

The property of cross products:

  $\begin{array}{l}\left(DF\right)\sin {31}^{\circ }=\left(7.9\right)\sin {41}^{\circ }\\ DF=\frac{7.9\sin {41}^{\circ }}{\sin {31}^{\circ }}\\ DF\approx 10.1\end{array}$

Therefore, $m\angle F\approx 31$ , $EF\approx 14.6$ and $DF\approx 10.1$ .

The triangle using law of sine:

  

(b)

To determine

To draw:

The triangle using the Law of Cosines

Answer to Problem 57HP

The triangle using law of Cosines:

  

Explanation of Solution

Given information:

Law of cosine:

  $a^2=b^2+c^2-2bc\cos A$

Interpretation:

If $\Delta ABC$ has length $a$ , $b$ and $c$ representing the length of sides opposite angles with measure $A$ , $B$ and $C$ :

Law of cosine:

  $a^2=b^2+c^2-2bc\cos A$

Simplify the equation:

  $\begin{array}{l}\cos Y=\frac{{190}^2+{184}^2-{75}^2}{2\left(190\right)\left(184\right)}\\ \cos Y=0.9201\\ m\angle Y={\cos }^{-1}0.9201\\ m\angle Y=23\end{array}$

Also, Law of cosines is:

  $\begin{array}{l}\cos X=\frac{{190}^2+{75}^2-{184}^2}{2\left(190\right)\left(75\right)}\\ \cos X=0.27611\\ m\angle X={\cos }^{-1}0.27611\\ m\angle X=74\end{array}$

The theorem of triangle angle sum:

  $\begin{array}{l}m\angle Z=180-m\angle X-m\angle Y\\ m\angle Z=180-23-74\\ m\angle Z=83\end{array}$

Therefore, $m\angle X\approx 74$ , $m\angle Y\approx 23$ and $m\angle Z\approx 83$ .

The triangle using law of Cosines: