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.
Find angle CAD=x For triangle ABC ,D is a point inside triangle.
geometry
Best Answer
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 $x=14^{\circ}$