Chapter 16.1, Problem 3MRE

To determine

To solve: the given triangle.

Answer to Problem 3MRE

The given triangle $c=3,\angle A=45°,\text{and}\angle B=45°$ .

Explanation of Solution

Given information:

Give lengths to three significant digits and angle measures to the nearest tenth of a degree.

  $a=\frac{3\sqrt{2}}{2},b=\frac{3\sqrt{2}}{2},\angle C=90°$

Calculation:

Summarize the information in a figure.

  

Since two sides and the included angle are known, the triangle is as an SAS triangle.

Use the law of cosine to find $a$ .

  $c^2=a^2+b^2-2ab\cos C$

Substitute $\frac{3\sqrt{2}}{2}$ for $b$ , $\frac{3\sqrt{2}}{2}$ for $a$ and $90°$ for $C$ , and simplify

  $\begin{array}{c}{c}^2={\left(\frac{3\sqrt{2}}{2}\right)}^2+{\left(\frac{3\sqrt{2}}{2}\right)}^2-2\left(\frac{3\sqrt{2}}{2}\right)\left(\frac{3\sqrt{2}}{2}\right)\cos 90°\\ =\left(\frac{18}{4}\right)+\left(\frac{18}{4}\right)-2\left(\frac{18}{4}\right)0\\ =\frac{36}{4}\\ =9\end{array}$

Take the positive square root on both the sides.

  $\begin{array}{c}\sqrt{c^2}=\sqrt{9}\\ c=3\end{array}$

Next, to find the angles.

Let us find angle $B$ using the law of sines.

  $\frac{\sin B}{b}=\frac{\sin C}{c}\text{or}\sin B=\frac{b\sin C}{c}$

Substitute $\frac{3\sqrt{2}}{2}$ for $b$ , 3 for $c$ and $90°$ for $C$ , and simplify

  $\begin{array}{c}\sin B=\frac{\frac{3\sqrt{2}}{2}\sin 90°}{3}\\ =\frac{\sqrt{2}}{2}\end{array}$

Use the inverse sine function to find $B$ .

  $\begin{array}{c}B={\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)\\ =45°\end{array}$

The sum of all the angles on a triangle is $180°$ .

Subtract $90°$ and $45°$ from $180°$ to find $A$ ,

  $180°-90°-45°=45°$

Therefore, for the given triangle $c=3,\angle A=45°,\text{and}\angle B=45°$ .