# Trigonometry: Isosceles Triangle [duplicate]

I saw the following problem on Facebook (figure not drawn to scale): $\triangle ABC$ is an isosceles triangle with $AB\equiv BC$, $AC\equiv BD$, and $\angle B=20^\circ$. Find $\angle CAD$.

What I did was let $BC\equiv x$ and divide the triangle in two by bisecting $\angle B$. This yielded two right triangles with base $x\cos80^\circ$. Therefore, the base of $\triangle ABC$ has length $2x\cos80^\circ$.

Clearly, $DC=x-2x\cos80^\circ$, and $\angle C=80^\circ$.

Now, to find $AD\equiv y$, I used the law of cosines:

$$y=\sqrt{(2x\cos80^\circ)^2+(x-2x\cos80^\circ)^2-2(2x\cos80^\circ)(x-2x\cos80^\circ)\cos80^\circ}.$$

Finally, to find $\angle CAD\equiv\theta$, I used the law of sines:

\begin{align} \frac{\sin80^\circ}{y}&=\frac{\sin\theta}{x-2x\cos80^\circ}\\ \Longrightarrow\theta&=70^\circ. \end{align}

Is my result correct? Is there an easier way to solve this problem?

## marked as duplicate by azimut, Thomas, Daniel Robert-Nicoud, Nick Peterson, Vedran ŠegoOct 29 '13 at 18:00

• For example, construct $E$ in the triangle such that $ACE$ is equilateral. Then $BE$ bisects angle $B$ and triangle $DBA$ is congruent to triangle $ECB$. Thus $\angle{CAD}=80^{\circ}-\angle{BAD}=80^{\circ}-\angle{CBE}=80^{\circ}- \frac{\angle{CBA}}{2}=70^{\circ}$ – Ivan Loh Oct 29 '13 at 17:07