Chapter 8.6, Problem 38PPS

To determine

To calculate:

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

Answer to Problem 38PPS

The measurements of triangle are $m\angle X\approx {80}^{\circ },m\angle Y\approx {58}^{\circ },m\angle Z\approx {42}^{\circ }$ .

Explanation of Solution

Given information:

  

Concept used:

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $a,b=$ sides

  $\cos C=$ angle

Calculation:

  

With the help of the above diagram it is easy to get the three sides of triangle and non-included angle. So, with the help of the Law of cosines get the missing angle.

First consider $Z$ to be the angle.

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $\begin{array}{l}a=43\\ b=50\\ c=34\\ \cos C=Z\end{array}$

First substitute,

  ${\left(34\right)}^2={\left(43\right)}^2+{\left(50\right)}^2-2\left(43\right)\left(50\right)\cos \left(Z\right)$

Now simplify,

  $1156=1849+2500-4300\cos Z$

Now solve for the $Z$

  $0.742558\approx \cos Z$

Now, get the value of $Z$ by using the inverse cosine ratio

  $Z\approx 42$

Similarly, use the law of cosine to get the measure of $Y$

First substitute,

  ${\left(43\right)}^2={\left(34\right)}^2+{\left(50\right)}^2-2\left(34\right)\left(50\right)\cos Y$

Now simplify,

  $1849=1156+2500-3400\cos Y$

Now solve for the $Y$

  $0.53147\approx \cos Y$

Now, get the value of $Y$ by using the inverse cosine ratio

  $Y\approx 58$

The sum of the angles of a triangle is $180$ . So,

  $m\angle X=180-\left(42+58\right)=80$

Hence, the measurements of triangle are $m\angle X\approx {80}^{\circ },m\angle Y\approx {58}^{\circ },m\angle Z\approx {42}^{\circ }$