Department of Mathematics


Worked Examples

Right-angle Triangles

Part 1: Trigonometric ratios

In a right-angle triangle, the side across from the right-angle is called the hypotenuse. Let $\theta$ be one of the other angles. The side across from that angle is called the opposite side to $\theta$ and the third side is called the adjacent side to $\theta$, as shown in the picture below.

We can compute the values of the trigonometric functions at $\theta$ using the ratios of the side lengths in the triangle as follows. $$\begin{alignat*}{3} \sin\left(\theta\right) &=\frac{\text{opposite}}{\text{hypotenuse}} \hspace{20 mm} &\csc\left(\theta\right) &= \frac{1}{\sin\left(\theta\right)}=\frac{\text{hypotenuse}}{\text{opposite}} \\ \\ \cos\left(\theta\right) &=\frac{\text{adjacent}}{\text{hypotenuse}} \hspace{20 mm} &\sec\left(\theta\right) & = \frac{1}{\cos\left(\theta\right)}=\frac{\text{hypotenuse}}{\text{adjacent}} \\ \\ \tan\left(\theta\right) &=\frac{\text{opposite}}{\text{adjacent}} &\cot\left(\theta\right) & =\frac{1}{\tan\left(\theta\right)}=\frac{\text{adjacent}}{\text{opposite}} \end{alignat*}$$

Part 2: Two special triangles

Consider a triangle whose angles are $45^{\circ}-45^{\circ}-90^{\circ}$. It will be similarsame shape but possibly different size to a triangle with side lengths $1, 1$ and $\sqrt{2}$, so it will have the same ratios for the side lengths. (Details)By the similarity, our triangle will have side lengths $c,c$ and $\sqrt{2}c$ for some positive number $c$. When taking the ratio of two side lengths, the factors of $c$ from both sides will cancel, e.g. $c/\sqrt{2}c=1/\sqrt{2}$.

Then, we get: $$\sin\left(45^{\circ}\right)=\frac{1}{\sqrt{2}}, \quad \cos\left(45^{\circ}\right)=\frac{1}{\sqrt{2}}, \quad \tan\left(45^{\circ}\right)=1$$
Another triangle worth remembering has angles $30^{\circ}-60^{\circ}-90^{\circ}$. It is similar to a triangle with side lengths $1, 2$ and $\sqrt{3}$, where the smallest side is across from the smallest angle and the biggest one--across from the biggest angle.

Then we get: \begin{align*} \sin\left(30^{\circ}\right)=\frac{1}{2} \hspace{20 mm} & \sin\left(60^{\circ}\right)=\frac{\sqrt{3}}{2} \\ \cos\left(30^{\circ}\right)=\frac{\sqrt{3}}{2} \hspace{20 mm} & \cos\left(60^{\circ}\right)=\frac{1}{2} \\ \tan\left(30^{\circ}\right)=\frac{1}{\sqrt{3}} \hspace{20 mm} & \tan\left(60^{\circ}\right)=\sqrt{3} \\ \end{align*}