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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.

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

1 Answer 1

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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 $x=14^{\circ}$

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