Chapter 6, Problem 1RE

In Exercises 1-12, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. If no triangle exists, state “no triangle.” If two triangles exist, solve each triangle.

$A={70}^{\circ },B={55}^{\circ },a=12$

To determine

To calculate: The length of triangle to the nearest tenth and angle measure to the nearest degree whose angles and side is $A=70°,B=55°,a=12$.

Answer to Problem 1RE

Solution: The missing values are: $C=55°,b\approx 10.4\text{ and }c\approx 10.5$

Explanation of Solution

Given Information: The values: $A=70°,B=55°,a=12$.

Formula used:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\mathrm{sinC}}$.

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

Calculation:

For any triangle,

The Law of sine states:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\mathrm{sinC}}$

Using the angle sum property we will obtain:

$\begin{array}{c}A+B+C=180°\\ 70°+55°+C=180°\\ C=180°-125°\\ C=55°\end{array}$

Using the law of sines we get,

$\begin{array}{c}\frac{a}{\sin A}=\frac{c}{\mathrm{sinC}}\\ \frac{12}{\sin 70°}=\frac{c}{\sin 55°}\\ c\cdot \sin 70°=12\cdot \sin 55°\\ c=\frac{12\cdot \sin 55°}{\sin 70°}\end{array}$

This gives $c\approx 10.5$ to the nearest tenth.

Also,

$\begin{array}{c}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{12}{\sin 70°}=\frac{b}{\sin 55°}\\ b\cdot \sin 70°=12\cdot \sin 55°\\ b=\frac{12\cdot \sin 55°}{\sin 70°}\end{array}$

This gives $b\approx 10.5$ to the nearest tenth.

So, $C=55°,b\approx 10.4\text{ and }c\approx 10.5$.

Hence, $C=55°,b\approx 10.4\text{ and }c\approx 10.5$.