Chapter 13.5, Problem 4CYU

To determine

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

Answer to Problem 4CYU

In triangle ABC

So side AC= b= 6.1

Angle $\angle BAC$ = ${42}^{\circ }$

And Angle $\angle BCA$ = ${28}^{\circ }$

Explanation of Solution

Given information:

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

AB= c = 3

BC=a=6

∠ABC= $\angle B$ = ${110}^{\circ }$

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 = 4

BC = a = 3

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow b^2=6^2+3^2-2\times 6\times 3\times \cos {110}^{\circ }\\ \Rightarrow b^2=36+9-36\times (-0.342)\\ \Rightarrow b^2=25+12.312\\ \Rightarrow b^2=37.312\\ \Rightarrow b=\sqrt{37.312}\\ \Rightarrow b=6.1\end{array}$

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}B}{AC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{3\times \mathrm{Sin}{110}^{\circ }}{6.1}\\ \Rightarrow \mathrm{Sin}C=\frac{3\times (0.9396)}{6.1}\\ \Rightarrow \mathrm{Sin}C=0.4621\\ \Rightarrow C={\sin }^{-1}(0.4621)\\ \Rightarrow C={27.53}^{\circ }\\ \Rightarrow C={28}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A={180}^{\circ }-(B+C)\\ \Rightarrow A={180}^{\circ }-({110}^{\circ }+{28}^{\circ })\\ \Rightarrow A={180}^{\circ }-{138}^{\circ }\\ \Rightarrow A={42}^{\circ }\end{array}$

In triangle ABC

So side AC= b= 6.1

Angle $\angle BAC$ = ${42}^{\circ }$

And Angle $\angle BCA$ = ${28}^{\circ }$