Chapter 13.5, Problem 17PPS

To determine

To find: To determine whether the each triangle is solved by cosines rule or sine rule in the beginning. Also have to find the other sides and angles.

Answer to Problem 17PPS

We will use sine rule in the beginning because one angle and its opposite side is given and there is other side given.

In triangle ABC

So side BC= a= 18.25

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

And Angle $\angle BCA=\angle C$ = ${45}^{\circ }$

Explanation of Solution

Given information:

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

AB= c = 13

AC= b =14

∠ABC= $\angle B$ = ${50}^{\circ }$

Calculation:

So we will use sine rule in the beginning because one angle and its opposite side is given and there is other side given.

To find the angle, we use the sine formula for any triangle:

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}B}{AC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{13\times \mathrm{Sin}{50}^{\circ }}{14}\\ \Rightarrow \mathrm{Sin}C=\frac{13\times (0.766)}{14}\\ \Rightarrow \mathrm{Sin}C=0.7112\\ \Rightarrow C={\sin }^{-1}(0.7112)\\ \Rightarrow C={45.33}^{\circ }\\ \Rightarrow C={45}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A={180}^{\circ }-(B+C)\\ \Rightarrow A={180}^{\circ }-({50}^{\circ }+{45}^{\circ })\\ \Rightarrow A={180}^{\circ }-{95}^{\circ }\\ \Rightarrow A={85}^{\circ }\end{array}$

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ \Rightarrow a^2={14}^2+{13}^2-2\times 14\times 13\times \cos {85}^{\circ }\\ \Rightarrow a^2=196+169-364\times (0.0871)\\ \Rightarrow a^2=365-31.72\\ \Rightarrow a^2=333.27\\ \Rightarrow a=\sqrt{333.27}\\ \Rightarrow a=18.25\end{array}$

In triangle ABC

So side BC= a= 18.25

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

And Angle $\angle BCA=\angle C$ = ${45}^{\circ }$