# Find the angle $\angle ACD=\theta$.

Problem: Triangle $\triangle ABC$ has $AC=BC$ and $\angle ACB=96^{\text{o}}$. $D$ is a point in $\triangle ABC$ such that $\angle DAB=18^{\text{o}}$ and $\angle DBA=30^{\text{o}}$. Compute $\angle ACD.$

Drawing the problem first: From the law of cosine we have that $b^2=2a^2-2a^2\cos{96}=2a^2(1-\cos{96})$ which leads to $b=|a|\sqrt{2(1-\cos{96})}.$

Since $\angle CAB=\angle ABC=(180-96)/2=42$, it follows that $\angle CAD=42-18=24$ and $\angle CBD = 42-30=12$. This information allows us to express $f$ in two ways by the law of cosine: \begin{array}{lcl} f^2 & = & e^2+a^2+2ea\cos{12} \ \ \ \ \ \ \ \ \ \quad \ \quad (1)\\ f^2 & = & d^2+a^2+2da\cos{24} \ \ \ \ \ \ \ \ \ \quad \ \quad (2)\\ \end{array} Subtracting (1)-(2) we get $e^2-d^2+2a(e\cos{12}-d\cos{24})=0$.

Here I'm stuck. Any suggestion on how to proceed?

• Well, we know $\angle CDA = 156 - \theta$, $\angle DCB = 96 - \theta$, $\angle ADB = 132$, and $\angle CDB = 180 - (96 - \theta + 12) = 72 + \theta$ Jun 29 '17 at 19:44

Using Law of Sines in triangle $\triangle{ABC}$ we can write: $$\frac{a}{\sin 42}=\frac{b}{\sin96} \Rightarrow b=\frac{a\sin 96}{\sin42}$$ Similarly in triangle $\triangle{ABD}$ we have: $$\frac{b}{\sin 132}=\frac{d}{\sin30} \Rightarrow b=2d\sin 132=2d\cos42$$ Equating $b$'s gives us: $$a = \frac{2d\sin 42 \cos 42}{\sin 96}=\frac{d \sin84}{\cos6}=\frac{d\cos6}{\cos6} \Rightarrow a=d$$ So triangle $\triangle{ACD}$ is an isosceles triangle which yields $\measuredangle{ACD}=78^{\circ}$

• This is a great solution. I hate that I always make my math so hard and the arithmetic so complex... Would it be possible to solve this problem with my method? Jun 29 '17 at 20:01
• Seems like we can't conclude a relationship between lengths from your last equation. But maybe others can proceed.
– user373239
Jun 29 '17 at 20:15

Let $CF$ be an altitude of $\Delta ABC$ and $DB\cap CF=\{E\}$.

Hence, $\measuredangle ECA=\frac{1}{2}\cdot96^{\circ}=48^{\circ}$ and $\measuredangle EDA=18^{\circ}+30^{\circ}=48^{\circ}$.

Now, since $F$ is a midpoint of $AB$, we obtain $AE=EB$, which says that $\measuredangle EAB=30^{\circ}$

and from here $\measuredangle EAD=30^{\circ}-18^{\circ}=12^{\circ}.$

But $\measuredangle CAB=90^{\circ}-48^{\circ}=42^{\circ},$ which gives $\measuredangle CAE=42^{\circ}-30^{\circ}=12^{\circ}$.

Thus, $\Delta CAE\cong\Delta DAE,$ which gives $CE=DE$ and since $\measuredangle CED=90^{\circ}+30^{\circ}=120^{\circ}$,

we obtain $\measuredangle ECD=30^{\circ}$ and $\measuredangle ACD=48^{\circ}+30^{\circ}=78^{\circ}$ and we are done!

• @Parseval What do you think about my solution? Thank you! Jun 30 '17 at 13:45