# Given the vertex angle and side lengths of an isosceles, find the base

I need to be able to do this programmatically, so I'll need to be able to convert an example into algebra, but for the sake of hopefully having it make more sense to me, let's say the two sides are 15 units long, and the angle of the vertex is 12 degrees. How would I go about determining the length of the base?

-

Hint:

• Sum of the angles in a triangle in $180^\circ$.

• In a triangle, angles opposite to the equal sides are equal.

• Sine Rule

• $\sin\left(\dfrac \pi 2-\alpha \right)=\cos \alpha$ and $\sin \alpha=2 \sin \dfrac\alpha 2 \cos \dfrac\alpha 2$

Let us consider the triangle $\Delta ABC$, such that $AB=AC=a$ and let $BC=b$. Now, we know that angles $\angle ABC=\angle ACB$ since they are the angles opposite to equal sides $AB$ and $AC$. Let's say $\angle ABC=\theta$ and you know the angle at the vertex, say $\alpha$. Now, applying the angle sum property of the triangles, you have $\alpha+2 \theta=180^\circ$. Now, apply the sine rule,$\dfrac{a}{\sin \theta}=\dfrac{b}{\sin \alpha}$. Recall that we'd like to know $b$: $b=\dfrac{a \sin \alpha}{\sin\left(\dfrac \pi 2-\dfrac \alpha 2\right)}=\dfrac{a\sin \alpha}{\cos \dfrac \alpha 2}=\dfrac{2a \sin \dfrac \alpha 2\cos \dfrac \alpha 2}{\cos \dfrac \alpha 2}=2a\sin \dfrac \alpha 2$

Alternatively, you can use the law of the cosines which would yield the answer in a single step.

-

I'd use the law of cosines.

-