geometrytriangles
Given that
ABC is a triangle, $\hat{ABD} = 30^\circ, \hat{ACD} = \hat{DBC} = 10^\circ, \hat{DCB} = 20^\circ, \hat{BAD} = a$
Evaluate $a$
How do we solve this triangle using Law of Sines? I've seen something like
$$\biggr (\dfrac{\sin(110-a)}{\sin a}\biggr )\biggr (\dfrac{\sin80}{\sin10}\biggr )=1$$
but I dont know if that's related to Law of Sines.
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*} AB&=BC\\ \implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ. \end{align*}
Erect an equilateral triangle $ACE$ on base $AC$. Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$
$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus $$\angle DEC=\angle DBC=20^\circ.$$
\begin{align*} \angle ECB&=\angle ECA-\angle BCA=10^\circ\\ \implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC. \end{align*}
Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus
\begin{align*} \angle CAD&=\angle DAE=30^\circ\\ \angle BAE&=\angle CAE-\angle CAB=10^\circ\\ \implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\ \implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ. \end{align*}
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 $x=14^{\circ}$
Best Answer
It should be immediate that $\angle BAC = 110^\circ $. The Law of Sines applied to the three small triangles (the ones with $D$ as a vertex) gives you the following equalities: $$ \frac{\sin a }{\overline{BD}} = \frac{\sin 30}{\overline{AD}} ~~~~~~~~ \frac{\sin 20 }{\overline{BD}} = \frac{\sin 10}{\overline{CD}} ~~~~~~~~ \frac{\sin 10 }{\overline{AD}} = \frac{\sin (110 - a)}{\overline{CD}} $$ It turns out that this is nice, because you can essentially cancel out the lengths of the sides. Rearrange the quotients as follows: $$ \frac{ \overline{AD}}{\overline{BD}} = \frac{\sin 30}{\sin a} ~~~~~~~~ \frac{\overline{BD}}{ \overline{CD}} = \frac{\sin 20}{\sin 10} ~~~~~~~~ \frac{\overline{CD}}{\overline{AD}} = \frac{\sin (110 - a)}{\sin 10 } $$ If you multiply all the left sides together and multiply all the right sides together, you get $$ 1 = \frac{\sin 30 \sin 20 \sin (110 - a)}{\sin a \sin 10 \sin 10}$$ From here, we'll be a little sneaky. First use $\sin 30 = \frac{1}{2}$ to get: $$ {\sin a \sin 10 \sin 10} = \frac{1}{2} \sin 20 \sin (110 - a)$$ Now use the double-angle identity for $\sin 20$ and convert $\sin a$ to a cosine to get: $$ \cos (a - 90) \sin 10 = \cos 10 \sin(110 - a)$$ From here, taking $\boxed{a = 100}$ solves the equation. Admittedly, this is a little hacky, but there may be a better way to get to the final answer after cancelling side lengths.