Sides of isosceles triangle I have an isosceles triangle; the vertex angle and the base length are known (I think the base angles can be figured from this). How do I calculate the leg length?
 A: Any isosceles triangle is symmetric about a line $\mathcal L$. Draw that line and determine the angles formed at the intersection of $\mathcal L$ and base, which can be done using the symmetry noted above. You should be able to see how go from here.
A: The law of sines would work here. If $\alpha$ is the vertex angle, then the base angles are each $\frac{\pi-\alpha}{2}$, since all three angles must add up to $\pi$. The law of sines then tells us that
$$\frac{\text{base length}}{\sin(\alpha)}=\frac{\text{leg length}}{\sin(\frac{\pi-\alpha}{2})}$$
so that
$$\text{leg length}=\text{base length}\cdot\frac{\sin(\frac{\pi-\alpha}{2})}{\sin(\alpha)}.$$
This could probably be simplified slightly more with some trigonometric identities.
A: Let $A$ be the vertex where the equal sides meet. Let the vertex angle be $\theta$. Let the other two vertices be $B$ and $C$, and let $BC$ have length $a$.  
Draw a line from $A$ to the midpoint $M$ of the base $BC$.  
The side $AB$ of our triangle has length we will call $c$, We want to compute $c$.
By looking at the right-angled $\triangle ABM$, we can see that
$$\frac{a/2}{c}=\sin(\theta/2).$$
Now we can solve for $c$. We get $c=\dfrac{a/2}{\sin(\theta/2)}$.
Another way: Let the base be $a$, and the vertex angle $\theta$. Let the equal sides of the triangle each have length $x$. Then by the Cosine Law,
$$a^2=x^2+x^2-2x^2\cos \theta.$$
Now it is easy to solve for $x$.
