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I am unable to find out the unknown angles for the following triangle which I attached with this post.

enter image description here

Angle BAD and angle BCD are the unknown angles need to be calculated. Given that lines AB=BC=CD and angle CDE = 108 degrees

From my calculations: angle ADC = 180 - 108 = 72 degrees (angles on a straight line)

angle BAD + angle BCD = 108 degrees (exterior angle of a triangle = sum of interior opposite angles)

I could not proceed any further beyond this. I thought line DB (median) is bisecting angle ADC since line DB is bisecting line AC (AB=BC) but this only happens in case of isosceles and equilateral triangles.

I am very much stuck here and seek your kind suggestions here. I am also suspecting something could be wrong in the diagram of the triangle or may be the unknown angles. Let me know where I am wrong. Thanks in advance.

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  • $\begingroup$ You can use latex. It will be better. $\endgroup$ – tatan Oct 14 '16 at 14:49
  • $\begingroup$ Angle CDE cannot be 108 degrees as angle CAD is 26.6 degrees and angle CDE must be 90 + angle CAD. Unless of course A, D and E are not on a line. $\endgroup$ – Jens Oct 14 '16 at 14:57
  • $\begingroup$ @Jens How did you find $\angle CAD=26.6°$ $\endgroup$ – tatan Oct 14 '16 at 15:04
  • $\begingroup$ @tatan The triangle is a 1, 2, $\sqrt 5$ triangle. $\endgroup$ – Jens Oct 14 '16 at 15:05
  • $\begingroup$ @Jens it will be better if you elaborate. $\endgroup$ – tatan Oct 14 '16 at 15:07
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Hint- Using Law of Sines-

$$\frac{CD}{\sin\angle A}=\frac{AC (=2CD)}{\sin\angle ADC(=72^{\circ})} \implies \sin\angle A=\frac{\sin72^{\circ}}{2}$$

Hope this helps!!

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  • $\begingroup$ Thanks, but is it possible to solve this question without using trigonometry concepts? I tried using geometry principles but in vain. $\endgroup$ – tantrik Oct 15 '16 at 8:37
  • $\begingroup$ @tantrik As you can see, the angles are irrational. Hence, I don't think you can do it without trigonometry. $\endgroup$ – tatan Oct 15 '16 at 8:46
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Can you use trigonometry?

By the law of sines, and letting $\angle CAD = \theta$,

$\frac{\sin \theta}{1} = \frac {\sin 72^{\circ}}{2}$

$\sin \theta = \frac 12\sin 72^{\circ}$

I'm pretty sure trigonometry is needed here because the answer is not "nice" (in degrees at least).

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  • $\begingroup$ I don't see a need of having two exactly similar answers. $\endgroup$ – tatan Oct 14 '16 at 15:08
  • $\begingroup$ @tatan Hadn't you deleted yours? $\endgroup$ – Deepak Oct 14 '16 at 15:13
  • $\begingroup$ No...its there till now.... $\endgroup$ – tatan Oct 14 '16 at 15:14
  • $\begingroup$ @tatan I am pretty sure I saw it was deleted a while back. $\endgroup$ – Deepak Oct 14 '16 at 15:14
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    $\begingroup$ @tatan Why are you still posting comments then? I am prepared to drop it. $\endgroup$ – Deepak Oct 14 '16 at 15:18

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