Trigonometry

Sine and Cosine Laws

We discussed trigonometric ratios in right-angle triangles in the first section. What about oblique triangles, i.e. those without a right angle? The sine and cosine laws presented below are trigonometric relations that hold for any triangle.

Let $ABC$ be any triangle, labeled as below:

Part 1: The Sine Law

We have the following relation among the sides and angles of any triangle: $$\frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}$$

Example. Given a triangle $ABC$ with $c=5$ cm, $A=45^{\circ}$ and $B=60^{\circ}$, find the length of $a$.

We know that $A+B+C=180^{\circ}$, so $C=180^{\circ}-45^{\circ}-60^{\circ}=75^{\circ}$.

Using the Sine Law, $\frac{\sin\left(A\right)}{a} = \frac{\sin\left(C\right)}{c}$, we get: \begin{align*} \frac{\sin\left(45^{\circ}\right)}{a} &= \frac{\sin\left(75^{\circ}\right)}{5} = \frac{\sqrt{3}+1}{10\sqrt{2}} \\ \Rightarrow \quad \frac{1}{a\sqrt{2}} &= \frac{\sqrt{3}+1}{10\sqrt{2}} \\ \Rightarrow \quad a &= \frac{10}{\sqrt{3}+1} \approx 3.66 \text{ cm} \end{align*}

Part 2: The Cosine Law

The Cosine Law gives the following relation among the side lengths and angles in any triangle: \begin{align*} a^2 &= b^2 + c^2 - 2bc \cos\left(A\right) \\ b^2 &= a^2 + c^2 - 2ac \cos\left(B\right) \\ c^2 &= a^2 + b^2 - 2ab \cos\left(C\right) \end{align*}

Example. Find the angles in a triangle with side lengths $3, 4$ and $5$.

Using the Cosine Law, we get: \begin{alignat*}{3} 3^2 &= 4^2 + 5^2 - 2(4)(5) \cos\left(A\right) & \quad \Rightarrow \quad \cos\left(A\right)=\frac{4}{5} \\ 4^2 &= 3^2 + 5^2 - 2(3)(5) \cos\left(B\right) & \quad \Rightarrow \quad \cos\left(B\right)=\frac{3}{5} \\ 5^2 &= 3^2 + 4^2 - 2(3)(4) \cos\left(C\right) & \quad \Rightarrow \quad \cos\left(C\right)=0 \end{alignat*} This tells us that: $$A = \arccos\left(\frac{4}{5}\right) \approx 36.9^{\circ} \quad \quad B = \arccos\left(\frac{3}{5}\right) \approx 53.1^{\circ} \quad \quad C = 90^{\circ}$$