Chapter 13.5, Problem 1CYU

To determine

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

Answer to Problem 1CYU

In triangle ABC

So side AC= b= 5.1

Angle $\angle BAC$ = ${36}^{\circ }$

And Angle $\angle BCA$ = ${52}^{\circ }$

Explanation of Solution

Given information:

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

AB= 4

BC=3

∠ABC= 92

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 = 4

BC = a = 3

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow b^2=3^2+4^2-2\times 4\times 3\times \cos {92}^{\circ }\\ \Rightarrow b^2=9+16-24\times (-0.0349)\\ \Rightarrow b^2=25+0.8376\\ \Rightarrow b^2=25.8376\\ \Rightarrow b=\sqrt{25.8376}\\ \Rightarrow b=5.08\end{array}$

Or b =5.1

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}B}{AC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times \mathrm{Sin}{92}^{\circ }}{5.1}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times \mathrm{Sin}{92}^{\circ }}{5.1}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times \mathrm{Sin}{92}^{\circ }}{5.1}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times 0.999}{5.1}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times 0.999}{5.1}\\ \Rightarrow \mathrm{Sin}C=0.784\\ \Rightarrow C={\sin }^{-1}(0.784)\\ \Rightarrow C={51.62}^{\circ }\\ \Rightarrow C={52}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A={180}^{\circ }-(B+C)\\ \Rightarrow A={180}^{\circ }-({92}^{\circ }+{52}^{\circ })\\ \Rightarrow A={180}^{\circ }-{144}^{\circ }\\ \Rightarrow A={36}^{\circ }\end{array}$

In triangle ABC

So side AC= b= 5.1

Angle $\angle BAC$ = ${36}^{\circ }$

And Angle $\angle BCA$ = ${52}^{\circ }$