Chapter 8.6, Problem 41PPS

To determine

To calculate:

The $△ABC$ for given measures.

Answer to Problem 41PPS

The measurements of triangle are $m\angle A\approx {35}^{\circ },AB\approx 7.5,BC\approx 9.8$

Explanation of Solution

Given information:

Measurements of $△ABC$ :

  $\begin{array}{l}m\angle B=119\\ m\angle C=26\\ CA=15\end{array}$

Concept used:

Law of sines:

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

Calculation:

The given measurements of $△ABC$ are $m\angle B=119,m\angle C=26,CA=15$ .

So, it is easy to get the two sides of triangle and non-included angle. So, with the help of the Law of sines get the missing value of $AB$ .

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

  $m\angle W=180-\left(119+26\right)=35$

Law of sines:

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

First substitute,

  $\frac{\sin 26}{AB}=\frac{\sin 119}{15}$

Now use the cross-product property,

  $AB\cdot \sin 119=15\cdot \sin 26$

Divide each side by $\sin 119$

  $AB=\frac{15\cdot \sin 26}{\sin 119}$

Now, get the value of $AB$

  $AB\approx 7.5$

Similarly, use the law of sines to get the measurement of $BC$

First substitute,

  $\frac{\sin 35}{BC}=\frac{\sin 119}{15}$

Now use the cross-product property,

  $BC\cdot \sin 119=15\cdot \sin 35$

Divide each side by $\sin 119$

  $BC=\frac{15\sin 35}{\sin 119}$

Now, get the value of $BC$

  $BC\approx 9.8$

Hence, the measurements of triangle are $m\angle A\approx {35}^{\circ },AB\approx 7.5,BC\approx 9.8$