# Week 17 – Precalc 11

One of the things we learned this week in Precalc 11 is how to use the cosine law to solve for angles or sidelengths of a triangle.

#### Cosine Law Formula

Sidelength Angle
$a^2=b^2+c^2-2bc CosA$ $Cos A=\frac{b^2+c^2-a^2}{2bc}$
$b^2=a^2+c^2-2ac Cos B$ $Cos B=\frac{a^2+c^2-b^2}{2ac}$
$c^2=a^2+b^2-2ab Cos C$ $Cos C=\frac{a^2+b^2-c^2}{2ab}$

#### Laws

Since there are two laws, Sine and Cosine. To determine which law to use, look at one of the angles if there is one given, if the opposite side has a given sidelength then use the Sine Law, if the opposite side has an unknown sidelength then use the Cosine Law.

#### Solving for an angle

Solve for ∠B.
First we need to figure out what formula to use.
Next, we can fill in the the values.
Then we can do the calculations.

#### Solving for a sidelength

Solve for side c.
First figure out what formula to use.
Next, fill in the values.
Then do the calculations.

#### Examples

In △ABC, AB=7, ∠B=105°, and BC=10; determine side b to one decimal place.

first we need to draw what the triangle might look like, it doesn’t have to be accurate.

$b^2=a^2+c^2-2ac Cos B$
$b^2=10^2+7^2-2(10)(7)Cos 105$°
$b^2=185.23$
$b=\sqrt{185.23}$
$b=13.6$

Solve for ∠Z to the nearest degree.

$Cos Z = \frac{x^2+y^2-z^2}{2xy}$
$Cos Z = \frac{23^2+11^2-14.7^2}{2(23)(11)}$
$Cos Z = \frac{433.91}{506}$
$Cos Z = 0.857$
$Z = Cos^{-1}(0.857)$
$Z = 31$°