contest-matheuclidean-geometrygeometrytrianglestrigonometry
$\triangle ABC$ has $AC=BC$, and $\angle ACB=96^\circ$. $D$ is a point in $\triangle ABC$, such that $\angle DAB=18^\circ, \angle DBA=30^\circ$. What is $\angle ACD$?
My attempt:
$$\angle ABC=\angle BAC=\frac{(180^\circ-96^\circ)}{2}=42^\circ.$$
$$\angle ADB=180^\circ-18^\circ-30^\circ=132^\circ.$$
From here onwards, I have no idea how to carry on. Can anyone help?
In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$
Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$
$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
Applying sine law in $\triangle ABC,$ $$\frac{AC}{\sin 42^\circ}=\frac{AB}{\sin 96^\circ}\implies \frac{AC}{AB}=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\cos 48^\circ}{2\sin 48^\circ \cos 48^\circ}$$ (applying $\sin 2A=2\sin A\cos A$)
So, $$AC=\frac{AB}{2\sin48^\circ} \implies AC=AD$$
So, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$
I don’t see why from $AC∥BE$, we get $∠CBA = (∠CBE) =\alpha$.
The way to do it is $\gamma = \angle ABD = \angle D = \angle ACB$ because ADBC is a //gm.
Let $\angle ABC = \theta$. From interior angle sums, we have $2 \gamma + \theta = 180^0$ and $\gamma + 2 \alpha = 180^0$
Result follows by eliminating $\gamma$.
Best Answer
When I see that $\angle CAB$ is partitioned as $24^\circ$ and $18^\circ$, what angle is left to make an $60^\circ$ angle (which is for making an equilateral triangle, which gives us more equal sides)? After doing that, we see that $\angle ECB = 36^\circ$ and this makes $\angle ABE = 30^\circ$. Now, notice that $\Delta ADB$ is congruent to $\Delta AEB$. So we have $|AD| = |AE| = |AC|$. So the answer is $78^\circ$.