# Laws of sines and cosines review

CCSS Math: HSG.SRT.D.11
Review the law of sines and the law of cosines, and use them to solve problems with any triangle.

## Law of sines

$\dfrac{a}{\sin(\alpha)}=\dfrac{b}{\sin(\beta)}=\dfrac{c}{\sin(\gamma)}$

## Law of cosines

$c^2=a^2+b^2-2ab\cos(\gamma)$

## Practice set 1: Solving triangles using the law of sines

This law is useful for finding a missing angle when given an angle and two sides, or for finding a missing side when given two angles and one side.

### Example 1: Finding a missing side

Let's find $AC$ in the following triangle:
According to the law of sines, $\dfrac{AB}{\sin(\angle C)}=\dfrac{AC}{\sin(\angle B)}$. Now we can plug the values and solve:
\begin{aligned} \dfrac{AB}{\sin(\angle C)}&=\dfrac{AC}{\sin(\angle B)} \\\\ \dfrac{5}{\sin(33^\circ)}&=\dfrac{AC}{\sin(67^\circ)}\\\\ \dfrac{5\sin(67^\circ)}{\sin(33^\circ)}&=AC \\\\ 8.45&\approx AC \end{aligned}

### Example 2: Finding a missing angle

Let's find $m\angle A$ in the following triangle:
According to the law of sines, $\dfrac{BC}{\sin(\angle A)}=\dfrac{AB}{\sin(\angle C)}$. Now we can plug the values and solve:
\begin{aligned} \dfrac{BC}{\sin(\angle A)}&=\dfrac{AB}{\sin(\angle C)} \\\\ \dfrac{11}{\sin(\angle A)}&=\dfrac{5}{\sin(25^\circ)} \\\\ 11\sin(25^\circ)&=5\sin(\angle A) \\\\ \dfrac{11\sin(25^\circ)}{5}&=\sin(\angle A) \end{aligned}
Evaluating using the calculator and rounding:
$m\angle A=\sin^{-1}\left(\dfrac{11\sin(25^\circ)}{5}\right)\approx 68.4^\circ$
Remember that if the missing angle is obtuse, we need to take $180^\circ$ and subtract what we got from the calculator.
Problem 1.1
$BC=$

Round to the nearest tenth.

Want to try more problems like this? Check out this exercise.

## Practice set 2: Solving triangles using the law of cosines

This law is mostly useful for finding an angle measure when given all side lengths. It's also useful for finding a missing side when given the other sides and one angle measure.

### Example 1: Finding an angle

Let's find $m\angle B$ in the following triangle:
According to the law of cosines:
$(AC)^2=(AB)^2+(BC)^2-2(AB)(BC)\cos(\angle B)$
Now we can plug the values and solve:
\begin{aligned} (5)^2&=(10)^2+(6)^2-2(10)(6)\cos(\angle B) \\\\ 25&=100+36-120\cos(\angle B) \\\\ 120\cos(\angle B)&=111 \\\\ \cos(\angle B)&=\dfrac{111}{120} \end{aligned}
Evaluating using the calculator and rounding:
$m\angle B=\cos^{-1}\left(\dfrac{111}{120}\right)\approx 22.33^\circ$

### Example 2: Finding a missing side

Let's find $AB$ in the following triangle:
According to the law of cosines:
$(AB)^2=(AC)^2+(BC)^2-2(AC)(BC)\cos(\angle C)$
Now we can plug the values and solve:
\begin{aligned} (AB)^2&=(5)^2+(16)^2-2(5)(16)\cos(61^\circ) \\\\ (AB)^2&=25+256-160\cos(61^\circ) \\\\ AB&=\sqrt{281-160\cos(61^\circ)} \\\\ AB&\approx 14.3 \end{aligned}
Problem 2.1
$m\angle A=$
$\Large{^\circ}$
Round to the nearest degree.

Want to try more problems like this? Check out this exercise.

## Practice set 3: General triangle word problems

Problem 3.1
"Only one remains." Ryan signals to his brother from his hiding place.
Matt nods in acknowledgement, spotting the last evil robot.
"$34$ degrees." Matt signals back, informing Ryan of the angle he observed between Ryan and the robot.
Ryan records this value on his diagram (shown below) and performs a calculation. Calibrating his laser cannon to the correct distance, he stands, aims, and fires.
To what distance did Ryan calibrate his laser cannon?
$\text{ m}$