Chapter 12.5, Problem 6CYU

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 6CYU

We will use cosine rule in the beginning because one angle is given and two sides are given making this angle.

In triangle ABC

In triangle ABC

So side AC= b= 6.7

Angle $\angle BAC$ = $\angle A$ = ${48}^{\circ }$

And Angle $\angle BCA$ = $\angle C$ = ${36}^{\circ }$

Explanation of Solution

Given information:

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

  AB= c = 4

  BC= a =5

  $\angle$ABC= $\angle B$ = ${96}^{\circ }$

Calculation:

We will use cosine rule in the beginning because one angle is given and two sides are given making this angle.

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

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow b^2=5^2+4^2-2\times 5\times 4\times \cos {96}^{\circ }\\ \Rightarrow b^2=25+16-40\times (-0.1045)\\ \Rightarrow b^2=41+4.1811\\ \Rightarrow b^2=45.1811\\ \Rightarrow b=\sqrt{45.1811}\\ \Rightarrow b=6.7\end{array}$

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}C}{AB}=\frac{\mathrm{Sin}B}{AC}\\ \Rightarrow \frac{\mathrm{Sin}C}{c}=\frac{\mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{c\times \mathrm{Sin}B}{b}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times \mathrm{Sin}{96}^{\circ }}{6.7}\\ \Rightarrow \mathrm{Sin}C=\frac{4\times 0.9945}{6.7}\\ \Rightarrow \mathrm{Sin}C=0.5937\\ \Rightarrow C={\sin }^{-1}(0.5937)\\ \Rightarrow C={36.42}^{\circ }\\ \Rightarrow C={36}^{\circ }\end{array}$

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow A={180}^{\circ }-(B+C)\\ \Rightarrow A={180}^{\circ }-({96}^{\circ }+{36}^{\circ })\\ \Rightarrow A={180}^{\circ }-{132}^{\circ }\\ \Rightarrow A={48}^{\circ }\end{array}$

In triangle ABC

So side AC= b= 6.7

Angle $\angle BAC$ = ${48}^{\circ }$

And Angle $\angle BCA$ = ${36}^{\circ }$