# Find angle CAD=x For triangle ABC ,D is a point inside triangle.

Find angle CAD=x For triangle ABC ,D is a point inside triangle such $$\angle ABD=30^{\circ}$$,$$\angle DBC=7^{\circ}$$,$$\angle ACD=\angle DCB=16^{\circ}$$ find messure of $$\angle CAD$$
Reflect B on CD to get E. then we know EBC=74 degrees and BA thus is the angle bisector of angle EBC. Also EDB=46 degree, but I think it is imposssible find that angle.

• Have you tried using the fact in all four triangles that the angles sum to $180^\circ$? Jun 3, 2020 at 18:43
• Yes, maby I can't see something. Jun 3, 2020 at 18:55

By Trigonometric Form of Ceva's Theorem,

$$\frac{\sin30^{\circ}}{\sin7^{\circ}}\frac{\sin16^{\circ}}{\sin16^{\circ}}\frac{\sin\widehat{CAD}}{\sin\widehat{BAD}}=1$$

Let $$\widehat{CAD}=x\,$$. Then the equation becomes

$$1=\frac{\sin30^{\circ}}{\sin7^{\circ}}\frac{\sin x}{\sin (111^{\circ}-x)}=\frac{1/2}{\sin7^{\circ}}\frac{\sin x}{\cos (21^{\circ}-x)}$$

Thus,

$$(1/2)[\sin(x-14^{\circ})+\sin(28^{\circ}-x)]=(1/2)(\sin x)$$

Equivalent to, $$\sin(x-14^{\circ})=\sin x-\sin(28^{\circ}-x)=2\cos14^{\circ}\sin(x-14^{\circ})$$

It satisfied only for $$x=14^{\circ}$$