0
$\begingroup$

enter image description here Can you please help me finding the value of angle $x$ in this image (I've drawn using microsoft paint, and added as many angles as many I could figure out). All angles are in degrees. Any exterior angle property or angle sum property seems not to help further.

$\endgroup$
  • $\begingroup$ What have you tried? What techniques are you supposed to be using? E.g. Are you allowed to use Ceva's Theorem? $\endgroup$ – Calvin Lin Apr 15 at 14:34
  • $\begingroup$ @CalvinLin, this is not a problem I've been given in my institution, neither it is related to my course. I was just trying it for fun, so any technique will do. $\endgroup$ – Martund Apr 15 at 14:35
  • $\begingroup$ Applying the trigonometric form of Ceva's theorem, we get that $ x = 78^\circ$. If you're interested, you can read up on it. (There will be other Euclidean geom / trigo approaches, but this is the fastest.) $\endgroup$ – Calvin Lin Apr 15 at 14:46
  • $\begingroup$ @CalvinLin, thank you, post it as an answer, I'll accept. $\endgroup$ – Martund Apr 15 at 15:06
  • $\begingroup$ I intentionally left it as a comment because I'd like to see an Euclidean geom approach which I think is "nicer". $\endgroup$ – Calvin Lin Apr 15 at 15:07
1
$\begingroup$

Draw equilateral triangle $ACE$ such that $E$ lies on the same side of $AC$ as $B$. Then angle chasing shows that $\angle EAB = 18^\circ = \angle BAD$. Since $AC=BC=EC$, we have that $\angle ABE =\frac 12 \angle ACE = 30^\circ = \angle DBA$. Hence triangles $ABE, ABD$ are congruent by ASA. Therefore $AE=AD$, but $AE=AC$, so $AD=AC$. From this we get $\angle ACD = 90^\circ - \frac 12 \angle DAC = 78^\circ$.

Sorry for not marking the angles on the figure. enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Since $AC=BC=EC$, we conclude that $C$ is the circumcircle of $\triangle ABE$. Hence, $\angle BAE=\dfrac12\angle BCE=30^\circ$, which leads to a contradiction apparently. $\endgroup$ – Martund Apr 15 at 17:42
  • $\begingroup$ @Martund You're making a mistake. In fact, $\angle BCE =36^\circ$ and there is no contradiction. $\endgroup$ – timon92 Apr 15 at 17:48
  • $\begingroup$ Thank You, I get it now. What software did you use to draw the image?? $\endgroup$ – Martund Apr 15 at 18:00
  • $\begingroup$ @Martund I used Geogebra. $\endgroup$ – timon92 Apr 15 at 18:04
2
$\begingroup$

Draw the angle bisector of $DAC$ and let it meet the $BD$ at $P$.
The rest is just angle chasing and your desire angles are $ACD=78°$ and $BCD=18°$ enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you show how you angle chase this? Thanks $\endgroup$ – timon92 Apr 15 at 16:58
  • $\begingroup$ @timon92, We observe that $\angle ABP=\angle BAP=30^\circ$, by construction. Hence, $P$ lies on angle bisector of $AB$ which passes through $C$, since $AC=BC$ is given. By angle sum property of $\triangle ABP$, we get $\angle APB=120^\circ$. By angle sum property of triangle $BPM$, where $M$ is the mid-point of $AB$, we get, $\angle BPM=60^\circ=\angle APM$. Hence, $\angle DPC=180^\circ-\angle BPM=120^\circ$. And$\angle APC=120^\circ$ by sum-of-angles-around-a-point property. $\triangle APD\cong\triangle APC$, by ASA rule. Hence, $DP=CP$. Angle sum property in $\triangle DPC$ will do now. $\endgroup$ – Martund Apr 15 at 17:53
  • $\begingroup$ @Martund Thanks, I see it now. I knew that some congruence should come up somewhere but I could not figure out where. Nice solution. $\endgroup$ – timon92 Apr 15 at 17:57
  • $\begingroup$ @Seyed, what software did you use to draw this figure?? $\endgroup$ – Martund Apr 15 at 18:01
  • $\begingroup$ @Martund, I used C.a.r. (Compass and Ruler) car.rene-grothmann.de/doc_en/index.html $\endgroup$ – Seyed Apr 15 at 19:04

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.