Triangles and Angles

Consider my diagram. In isosceles $\triangle ABD$ , let $\theta = \angle BAD=\angle BDA$ . Since $\angle BDA$ and $\angle ADC$ are supplementary angles (their sum is 180 degrees), $\angle ADC=180 -\theta$ .

Consider $\triangle ADC$ . Let $\angle DAC=\beta$ and $\angle DCB=\phi$ . We wish to find $\beta$ .

$\angle ADC=180 - \beta - \phi$

Equate: $\angle ADC=\angle ADC$

$180 -\theta\ = 180 - \beta - \phi$

$\theta = \beta + \phi$

From the problem, it says that

$\theta + \beta = 30 + \phi$

Now substitute $\theta = \beta + \phi$ to the above equation. We have

$\beta + \phi + \beta = 30 + \phi$

$2\beta = 30$

$\boxed{\beta = 15^\circ}$