$\triangle ABC$ 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, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$


enter image description here

  • 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;

enter image description here

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


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)


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

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

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

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


Have you drawn your picture?

I drew the picture out and I got an isosceles triangle, with base angles being ∠A and ∠B. Then I placed point D inside the triangle and connected that point to the 3 vertices of the triangle.

The first statement tells us that we are in an isosceles triangle ( meaning base angles are the congruent ). Therefore we can find the measure of the base angles since we know that ∠ACB=96∘. Each base angle turns out to be 42∘ each.

Consider triangle ABD. Can you find all angles of that triangle? And can you then find the angles of triangle BDC? and ADC?

I hope this helped! If you are still stuck let us know!

  • $\begingroup$ this probably should be a comment as it makes no attempt to answer the question at all. $\endgroup$ – user31280 Nov 28 '12 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.