Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I bet this question has been asked a million times, but I can't find a straight answer. I need to find the length of the hypotenuse in a triangle where I have one side and all the angles.


Triangle with one side and three angles

Now in the above triangle I have the length of a = 20 and all the angles. How do I - from here - get the length of the hypotenuse (c)?

share|cite|improve this question
according to the formula $c\cdot\cos 30 = a$ – W_D Sep 4 '13 at 12:41
up vote 4 down vote accepted

just use Law of sines ( it states that

$$\frac{a}{\sin{\alpha}} = \frac{b}{\sin{\beta}} = \frac{c}{\sin{\gamma}}$$

where $\alpha, \beta, \gamma$ are the angles opposited to sides $a, b, c$ respectively. Since $\gamma$ is a right angle, $\sin{\gamma} = 1$, and therefore in your example $c = \frac{a}{\sin{60°}}$.

share|cite|improve this answer
Thanks a lot. Exactly what I needed. Who knew it would be so simple :) – Trenskow Sep 4 '13 at 13:05
This is shooting an ant with a cannon! Just use the right-angle definitions of $\sin$ and $\cos$, as an earlier comment suggested. $a/c = \text{opposite/hypotenuse}=\sin 60^\circ$. – Ted Shifrin Sep 4 '13 at 15:02
if you have a nail to pe put on a wall, everything is a hammer :-) – mau Sep 5 '13 at 7:04

Maybe for that problem the short path is seeing the triangle as the half of an equilateral triangle. Therefore $20$ is the height and you know the relationship between the size and the height:

$$h=s\frac{\sqrt{3}}{2} \implies s=\frac{2h}{\sqrt{3}}=\frac{40}{\sqrt{3}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.