Chapter 13.5, Problem 9PPS

To determine

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

Answer to Problem 9PPS

In triangle ABC

So side AB= c= 3

Angle $\angle BAC$ = $\angle A$ = ${70}^{\circ }$

And Angle $\angle ABC=\angle B$ = ${40}^{\circ }$

Explanation of Solution

Given information:

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

BC= a = 3

AC= b =2

∠BCA=∠C= ${70}^{\circ }$

Calculation:

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

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos C\\ \Rightarrow c^2=3^2+2^2-2\times 3\times 2\times \cos {70}^{\circ }\\ \Rightarrow c^2=9+4-12\times (0.342)\\ \Rightarrow c^2=13-4.104\\ \Rightarrow c^2=8.896\\ \Rightarrow c=\sqrt{8.896}\\ \Rightarrow c=2.98\end{array}$

Or c= 3

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}A}{BC}=\frac{\mathrm{Sin}C}{c}\\ \Rightarrow \frac{\mathrm{Sin}A}{a}=\frac{\mathrm{Sin}C}{c}\\ \Rightarrow \mathrm{Sin}A=\frac{3\times \mathrm{Sin}{70}^{\circ }}{3}\\ \Rightarrow \mathrm{Sin}A=\frac{3\times {0.9396}^{}}{3}\\ \Rightarrow \mathrm{Sin}A=0.9396\\ \Rightarrow A={\sin }^{-1}(0.9396)\\ \Rightarrow A={70}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow B={180}^{\circ }-(A+C)\\ \Rightarrow B={180}^{\circ }-({70}^{\circ }+{70}^{\circ })\\ \Rightarrow B={180}^{\circ }-{140}^{\circ }\\ \Rightarrow B={40}^{\circ }\end{array}$

In triangle ABC

So side AB= c= 3

Angle $\angle BAC$ = $\angle A$ = ${70}^{\circ }$

And Angle $\angle ABC=\angle B$ = ${40}^{\circ }$