Can the vertex angle of an isosceles triangle be found without the law of cosines (no calculator)?

Question

If we know three sides of an isosceles triangle, can we find the measure of the angles without using a calculator (that means no law of Cosines/Sines).

Answer 1

Call the sides of the triangle lengths $a$, $a$, and $b$. Drawing a line of symmetry through the triangle lets us divide the triangle into two right triangles. Hence we can calculate $$\cos(\theta)=\frac{\frac{b}{2}}{a}=\frac{b}{2a}$$ where $\theta$ is the repeated angle in the triangle. This tells us the angles of the triangle are $\theta$, $\theta$, and $90-2\theta$, where $$\theta=\cos^{-1}(\frac{b}{2a}).$$

Answer 2

Yes. We can use the law of sines.

That is, choose two sides of unequal lengths $a$ and $b$. Let $\alpha$ and $\beta$ denote the measures of their respective opposite angles$. Then we have:

$$\frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b}$$

If $a < b$, then $2\alpha + \beta = 180^\circ$ since the triangle is isosceles. From here, we have $\beta = 180^\circ - 2\alpha$. Plug this into the above, and we have an equation with a single unknown.