# Find an angle in a given triangle

$\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?

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, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$ • Extend $AD$ to meet $CB$ at $M$.
• Draw a circle passing though $A$, $C$, and $M$ and cutting $AB$ at $L.$
• Draw the segment $ML$, let the segment $CL$ cut $AM$ and $BD$ at $P$ and $Q$ respectively.
• $ACML$ is a cyclic quadrilateral, thus $L\hat MB=C\hat AL=42° \implies$ $\triangle MLB$ is isosceles since $M\hat BL=42°$ with $LB=LM$
• The inscribed angle theorem implies; \begin{align}&1.\ \ C\hat LM=C\hat AM=24° \implies C\hat LA=60° \text{ ( angles on a straight line.)}\\&2.\ \ C\hat M A=C\hat LA=60°\implies A\hat ML=78° \text{ ( angles on a straight line.)}\end{align}
• $P\hat LM=24° \text { (since }M\hat LB=96°) \text{ ( angles on a straight line.)} \implies L\hat PM=78°$ and thus, $\triangle LPM$ is isosceles with $LP=LM$
• Using the $\text{ the sum of angles in a triangle}$ it is obvious that $L\hat QB =30°$ and thus $\triangle LQB$ is isosceles with $LQ=LB$
• We can then conclude that $LQ=LB=LM=LP$ but this can only hold if $P$ and $Q$ coincide since they're collinear. Hence,they coincide at point $D$, since $P$ moves on $DM$ and $Q$ moves on $BD$ and we have the diagram below; • Finally, by the inscribed angle theorem $A\hat CD=A\hat CL= A\hat ML=78°$

Thanks to Animoku Abdulwahab

$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$

As $\angle DAB=18^\circ, \angle DAC=(42-18)^\circ=24^\circ,$

Similarly, $\angle CBD=18^\circ$

Let $\angle ACD=x,$ so, $\angle DCB=96^\circ-x$

So, in $\triangle ADC, \angle ADC=180^\circ-(24^\circ+x)=156^\circ-x$

Similarly, from $\triangle BCD, \angle BDC=72^\circ-x$

Applying sine law in $\triangle BCD,$ $$\frac {CD}{\sin 12^\circ}=\frac {BC}{\sin(72^\circ+x)}$$

Similarly, from $\triangle ADC,$ $$\frac {CD}{\sin 24^\circ}=\frac {AC}{\sin(156^\circ-x)}$$

On division, $$\frac{\sin 24^\circ}{\sin 12^\circ}=\frac{\sin(156^\circ-x)}{\sin(72^\circ+x)}$$ as $AC=BC$

But $\sin(156^\circ-x)=\sin(180^\circ-(24^\circ+x))=\sin(24^\circ+x)$ and $\sin 24^\circ=2\sin 12^\circ\cos 12^\circ$

So, $$2\cos 12^\circ=\frac{\sin(24^\circ+x)}{\sin(72^\circ+x)}$$

Applying $2\sin A\cos B=\sin(A+B)+\sin(A-B),$

$$\sin(84^\circ+x)+\sin(60^\circ+x)=\sin(24^\circ+x)$$

or applying $\sin(A+B)$ formula and separating sine and cosine, $$\sin x (\cos24^\circ-\cos84^\circ-\cos60^\circ)=\cos x(\sin84^\circ-\sin24^\circ+\sin60^\circ)$$

$$\tan x=\frac{\sin84^\circ-\sin24^\circ+\sin60^\circ}{\cos24^\circ-\cos84^\circ-\cos60^\circ}=\frac{2\sin30^\circ\cos54^\circ+\sin60^\circ}{2\sin30^\circ\sin54^\circ-\cos60^\circ}$$ (applying $\sin C-\sin D,\cos C-\cos D$ formula)

$$=\frac{\cos54^\circ+\cos30^\circ}{\sin54^\circ-\sin30^\circ}=\frac{2\cos42^\circ\cos12^\circ}{2\cos42^\circ\sin12^\circ}$$

(applying $\sin C-\sin D,\cos C+\cos D$ formula)

$$=\cot 12 ^\circ=\tan(90-12)^\circ$$

$\implies x=78^\circ$ as $0<x<180^\circ$

• I started doing something similar too, I wonder if there is a simple purely geometric argument – Jean-Sébastien Nov 28 '12 at 14:05
• @Jean-Sébastien, please find my another answer. – lab bhattacharjee Nov 28 '12 at 14:11

Hint: