Chapter 13.5, Problem 10PPS

To determine

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

Answer to Problem 10PPS

In triangle ABC

So side AC= b= 19

Angle ∠BAC=∠A = ${48}^{\circ }$

And Angle ∠BCA=∠C = ${40}^{\circ }$

Explanation of Solution

Given information:

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

AB= c = 12

BC=a=14

∠ABC=∠B= ${92}^{\circ }$

Calculation:

To find the length of side AC, we use the cosine formula for any triangle:

  $b^2=a^2+c^2-2ac\cos B$

Let AC=b

AB = c = 12

BC = a = 14

  $\begin{array}{l}{b}^2={14}^2+{12}^2-2\times 14\times 12\cos {92}^{\circ }\\ \Rightarrow b^2={196}^{}+{144}^{}-2\times 4\times 3\times \cos {92}^{\circ }\\ \Rightarrow b^2=340-336\times (-0.0345)\\ \Rightarrow b^2=340+11.592\\ \Rightarrow b^2=351.592\\ \Rightarrow b=\sqrt{351.592}\\ \Rightarrow b=18.75\\ \Rightarrow b=19\end{array}$

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}B}{AC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{12\times \mathrm{Sin}{92}^{\circ }}{19}\\ \Rightarrow \mathrm{Sin}C=\frac{12\times 0.9993}{19}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times 0.999}{5.1}\\ \Rightarrow \mathrm{Sin}C=0.6311\\ \Rightarrow C={\sin }^{-1}(0.6311)\\ \Rightarrow C={39.13}^{\circ }\\ \Rightarrow C={40}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A={180}^{\circ }-(B+C)\\ \Rightarrow A={180}^{\circ }-({92}^{\circ }+{40}^{\circ })\\ \Rightarrow A={180}^{\circ }-{132}^{\circ }\\ \Rightarrow A={48}^{\circ }\end{array}$

In triangle ABC

So side AC= b= 19

Angle ∠BAC=∠A = ${48}^{\circ }$

And Angle ∠BCA=∠C = ${40}^{\circ }$