# Why is the unit circle and the trig functions defined on it useful, even when the hypotenuses of triangles in the problem are not 1?

Dec 12, 2014

Trig functions tell us the relationship between angles and side lengths in right triangles. The reason that they are useful has to do with the properties of similar triangles. Similar triangles are triangles that have the same angle measures. As a result, the ratios between similar sides of two triangles are the same for each side. In the image below, that ratio is $2$. The unit circle gives us relationships between the lengths of the sides of different right triangles and their angles. All of these triangles have a hypotenuse of $1$, the radius of the unit circle. Their sine and cosine values are the lengths of the legs of these triangles. Lets suppose we have a ${30}^{o}$- ${60}^{o}$- ${90}^{o}$ triangle and we know that the length of the hypotenuse is $2$. We can find a ${30}^{o}$- ${60}^{o}$- ${90}^{o}$ triangle on the unit circle. Since the hypotenuse of our new triangle is $2$, we know that the ratio of the sides is equal to the ratio of the hypotenuses.

$r = \frac{h y p o t e n u s e}{1} = \frac{2}{1} = 2$

So to solve the other sides of the triangle, we just need to multiply $\sin \left({30}^{o}\right)$ and $\cos \left({30}^{o}\right)$ by $r$, which is $2$.

$2 \sin \left({30}^{o}\right) = 2 \left(\frac{1}{2}\right) = 1$
$2 \cos \left({30}^{o}\right) = 2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$ You can solve any right triangle that you know at least one side of by finding a similar triangle on the unit circle, then multiplying $\sin \left(\theta\right)$ and $\cos \left(\theta\right)$ by the scaling ratio.