Chapter 12.5, Problem 14PPS

To determine

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

Answer to Problem 14PPS

side AB= c= 8.9

Angle $\angle$ABC=$\angle$B= ${15}^{\circ }$

And Angle $\angle$BAC=$\angle$A= ${85}^{\circ }$

Explanation of Solution

Given information:

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

  BC= a = 9

  AC=b=2

  $\angle$BCA=$\angle$C= ${80}^{\circ }$

Calculation:

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

Also as we know that-

  $\begin{array}{l}{c}^2=b^2+c^2-2bc\cos C\\ \Rightarrow c^2=9^2+2^2-2\times 9\times 2\times \cos {80}^{\circ }\\ \Rightarrow c^2=81+4-36\times (0.1736)\\ \Rightarrow c^2=85-6.2496\\ \Rightarrow c^2=78.75\\ \Rightarrow c=\sqrt{78}\\ \Rightarrow c=8.8741\\ \Rightarrow c=8.9\end{array}$

  $\begin{array}{l}\frac{\mathrm{Sin}A}{BC}=\frac{\mathrm{Sin}C}{AB}\\ \Rightarrow \frac{\mathrm{Sin}A}{a}=\frac{\mathrm{Sin}C}{c}\\ \Rightarrow \mathrm{Sin}A=\frac{a\times \mathrm{Sin}C}{c}\\ \Rightarrow \mathrm{Sin}A=\frac{9\times \mathrm{Sin}{80}^{\circ }}{8.9}\\ \Rightarrow \mathrm{Sin}A=\frac{9\times 0.9848}{8.9}\\ \Rightarrow \mathrm{Sin}A=0.9958\\ \Rightarrow A={\sin }^{-1}(0.9958)\\ \Rightarrow A={84.78}^{\circ }\\ \Rightarrow A={85}^{\circ }\\ \end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow B={180}^{\circ }-(A+C)\\ \Rightarrow B={180}^{\circ }-({85}^{\circ }+{80}^{\circ })\\ \Rightarrow B={180}^{\circ }-{165}^{\circ }\\ \Rightarrow B={15}^{\circ }\\ \end{array}$

In triangle ABC

So side AB= c= 8.9

Angle $\angle$ABC=$\angle$B= ${15}^{\circ }$

And Angle $\angle$BAC=$\angle$A= ${85}^{\circ }$