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$\triangle ABC$ 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 $\angle ACD$?

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4 Answers

up vote 3 down vote accepted

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

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THANKS for the answer –  chndn Nov 29 '12 at 10:27
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$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$

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 $0<x<180^\circ$

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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
    
THANKS for your help –  chndn Nov 29 '12 at 10:26
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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

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Hint:

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!

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