How to find the angle here?

Question

$ABC$ is an isosceles triangle. $AB=AC$ and $D$ is a point in the triangle. $DBC=25, DCB=10, CAB=110$. How much is the angle $ADB$?

Answer 1

After some angle-chasing, we repeatedly use the sine rule in various triangles. Since: $$\angle ABC=\angle ACB=35^\circ,$$ $$\angle ABD=10^\circ,$$ and $$\angle ACD=25^\circ;$$ So by using the sine rule: $$\sin 10^\circ=\frac{AD}{AB}\sin\angle ADB.$$ Similarly, $$\sin 25^\circ=\frac{AD}{AC}\sin\angle ADC,$$ and $$AB=AC\implies\frac{CD}{BD}=\frac{\sin 25^\circ}{\sin 10^\circ}=\frac{\sin\angle ADC}{\sin\angle ADB}.$$ But we also have $$\angle ADB+\angle ADC=215^\circ,$$ so $\angle ADB\approx 160.44^\circ$ as per @Dr.Mathva's comment.