Chapter 12.5, Problem 13PPS

To determine

To find: To round side lengths to the nearest tenth and angles to the nearest degree.

Answer to Problem 13PPS

In triangle ABC

So side AB= a= 6.9

Angle $\angle$ABC=$\angle$B= ${41}^{\circ }$

And Angle $\angle$BCA=$\angle$C= ${23}^{\circ }$

Explanation of Solution

Given information:

The length of two sides of triangle ABC and one angle.

  AB= c = 3

  AC=b=5

  $\angle$BAC=$\angle$A= ${116}^{\circ }$

Calculation:

To find the length of side BC= a , we use the cosine formula for any triangle:

Also as we know that-

  $\begin{array}{l}{a}^2=b^2+c^2-2ac\cos B\\ \Rightarrow a^2=5^2+3^2-2\times 5\times 3\times \cos {116}^{\circ }\\ \Rightarrow a^2=25+9-30\times (\text{-0}\text{.4383})\\ \Rightarrow a^2=34+13.149\\ \Rightarrow a^2=47.149\\ \Rightarrow a=\sqrt{47.149}\\ \Rightarrow a=6.86\\ \Rightarrow a=6.9\end{array}$

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}A}{BC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}A}{a}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}A}{a}\\ \Rightarrow \mathrm{Sin}C=\frac{3\times \mathrm{Sin}{116}^{\circ }}{6.9}\\ \Rightarrow \mathrm{Sin}C=\frac{3\times 0.8987}{6.9}\\ \Rightarrow \mathrm{Sin}C=0.3907\\ \Rightarrow C={\sin }^{-1}(0.3907)\\ \Rightarrow C={23.0}^{\circ }\\ \Rightarrow C={23}^{\circ }\\ \end{array}$

Since the sum of the angles of any triangle is ${180}^{\circ }$

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow B={180}^{\circ }-(A+C)\\ \Rightarrow B={180}^{\circ }-({116}^{\circ }+{23}^{\circ })\\ \Rightarrow B={180}^{\circ }-{139}^{\circ }\\ \Rightarrow B={41}^{\circ }\\ \end{array}$

In triangle ABC

So side AB= a= 6.9

Angle $\angle$ABC=$\angle$B= ${41}^{\circ }$

And Angle $\angle$BCA=$\angle$C= ${23}^{\circ }$