Chapter 8.6, Problem 9CYU

To determine

To calculate:

The each triangle

Answer to Problem 9CYU

The angles are $m\angle M={27}^{\circ },m\angle N={42}^{\circ },MP\approx 35.8$ .

Explanation of Solution

Given information:

  

Calculation:

  

Since, the given two sides and non included angles, setup the proportion using the Law of Sines to solve for measure of $\angle M$ :

Law of Sines:

  $a=b\frac{\sin \alpha }{\sin \beta }\mathrm{...}\left(1\right)$

Put the values in equation (1)

  $\frac{\sin M}{24.5}=\frac{\sin 11}{50}$

Apply the formula of cross product:

  $50\cdot \sin M=24.5\cdot \sin 111$

Divide both the sides by $\sin 50$

  $\begin{array}{l}\sin M=\frac{24.5\sin 111}{50}\\ m\angle M=27\end{array}$

The sum of angles of the triangle:

  $m\angle N=180-\left(111+27\right)=42$

Now, two sides and included angle, so use the Law of Cosines:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $\begin{array}{l}a=50\\ b=24.5\\ C=42\\ c=MP\end{array}$

Apply the formula of cross product:

  $\begin{array}{l}{\left(MP\right)}^2={\left(50\right)}^2+{\left(24.5\right)}^2-2\left(50\right)\left(24.5\right)\cos \left(42\right)\\ {\left(MP\right)}^2=1279.55\\ MP=\sqrt{1279.55}\\ MP\approx 35.8\end{array}$

The angles are $m\angle M={27}^{\circ },m\angle N={42}^{\circ },MP\approx 35.8$ .