Chapter 8.6, Problem 34PPS

To determine

To calculate:

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

Answer to Problem 34PPS

The measurements of triangle are $m\angle M\approx {19}^{\circ },m\angle N\approx {28}^{\circ },MN\approx 48.8$ .

Explanation of Solution

Given information:

  

Concept used:

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Law of sines:

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

Calculation:

  

With the help of the above diagram it is easy to get the two sides and an included angle. So, with the help of the Law of cosines get the value of $MN$

Use the law of cosines:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

So, first substitute

  $M{N}^2={\left(31.5\right)}^2+{\left(21.5\right)}^2-2\left(31.5\right)\left(21.5\right)\cos 133$

Now, simplify

  $M{N}^2=992.25+462.25+923.77$

Now, combine

  $M{N}^2=\sqrt{2378.27}$

Now get the square root value,

  $MN\approx 48.8$

Since, having the two sides and nonincluded angle, use the law of sines to get the measure of angle $M$

So, substitute

  $\frac{\sin 133}{48.8}=\frac{\sin M}{21.5}$

Now, cross products property

  $\sin M\cdot 48.8=\sin 133\cdot 21.5$

Now, solve for $M$ ,

  $\sin M=\frac{\sin 133\cdot 21.5}{48.8}$

Now use the inverse cosine ratio to get the value of $M$ ,

  $M\approx 19$

The sum of the angles of a triangle is $180$ . So, $m\angle N=180-\left(133+19\right)=28$

Therefore, the measurements of triangle are $m\angle M\approx {19}^{\circ },m\angle N\approx {28}^{\circ },MN\approx 48.8$