Chapter 14.5, Problem 31PPSE

Solution

To find: The remaining sides and angles of the given triangle.

  $A={56.1}^{\circ },B={63.8}^{\circ },C={60.1}^{\circ }$ .

Given:

  

Concept used:

Law of Cosines:

In a triangle ABC, let $a,b$ , and $c$ represents the length of the sides opposite $\angle A,\angle B$ , and $\angle C$ , respectively. Then,

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$ .

Calculation:

Here, $a=3.8,b=4.3,c=3.7$ .

The measure of $m\angle A$ can be found by using the Law of Cosines,

  $a^2=b^2+c^2-2bc\cos A$

By substituting the given values, it follows that

  $\begin{array}{c}{\left(3.8\right)}^2={\left(4.3\right)}^2+{\left(3.7\right)}^2-2\left(4.3\right)\left(3.7\right)\cos A\\ 14.44=18.49+13.69-31.82\cos A\\ 31.82\cos A=17.74\\ \cos A=0.5575\\ A={\cos }^{-1}\left(0.5575\right)\\ A\approx {56.1}^{\circ }\end{array}$

The measure of $m\angle B$ can be found by using the Law of Cosines,

  $b^2=a^2+c^2-2ac\cos B$

By substituting the given values, it follows that

  $\begin{array}{c}{\left(4.3\right)}^2={\left(3.8\right)}^2+{\left(3.7\right)}^2-2\left(3.8\right)\left(3.7\right)\cos B\\ 18.49=14.44+13.69-28.12\cos B\\ 28.12\cos B=9.64\\ \cos B=0.4417\\ B={\cos }^{-1}\left(0.4417\right)\\ B\approx {63.8}^{\circ }\end{array}$

Now since the sum of all the angles of a triangle is ${180}^{\circ }$ , therefore

  $\begin{array}{c}m\angle A+m\angle B+m\angle C={180}^{\circ }\\ {56.1}^{\circ }+{63.8}^{\circ }+m\angle C={180}^{\circ }\\ m\angle C={60.1}^{\circ }\end{array}$

Hence, the angles of the given triangle are $A={56.1}^{\circ },B={63.8}^{\circ },C={60.1}^{\circ }$ .