Chapter 8.6, Problem 37PPS

To determine

To calculate:

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

Answer to Problem 37PPS

The measurements of the triangle are $m\angle P\approx {35}^{\circ },m\angle R\approx {75}^{\circ },RP\approx 14.6$ .

Explanation of Solution

Given information:

  

Concept used:

Law of sines:

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

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

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 solve $PR$ .

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

First, substitute

  ${\left(PR\right)}^2=9^2+{15}^2-2\left(9\right)\left(15\right)\cos \left(70\right)$

Simplify,

  ${\left(PR\right)}^2=213.65$

Now take the square root of each side,

  $PR=\sqrt{216.65}$

The value of $PR$ is,

  $PR\approx 14.6$

Now, that having the two sides and a nonincluded angle, use the law of sines to get the measure of angle $R$ .

Law of sines:

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

Now, substitute

  $\frac{\sin 70}{14.6}=\frac{\sin R}{15}$

Use the cross products property

  $14.6\cdot \sin R=15\cdot \sin 70$

Now, divide each side by the $14.6$

  $\sin R=\frac{15\cdot \sin 70}{14.6}$

Use the inverse sine ratio

  $R\approx 75$

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

  $m\angle P=180-\left(75+70\right)=35$

Hence, the measurements of the triangle are $m\angle P\approx {35}^{\circ },m\angle R\approx {75}^{\circ },RP\approx 14.6$