# 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

$c^2=a^2+b^2-2ab\cos(\gamma)$
Want to learn more about the law of sines? Check out this video.
Want to learn more about the law of cosines? Check out this video.

## 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?
Do not round during your calculations. Round your final answer to the nearest meter.
$\text{ m}$
Want to try more problems like this? Check out this exercise.