# Is there a way to solve for the missing angle?

I was working on this problem. I tried to draw ${AC}$, ${BD}$ as isosceles triangle and divide into cases to find the missing angle $??$, but I got stuck. Can someone help me please or give me a clue?

• The smallest triangle on CD is isosceles, by the sine rule on triangles BCD and ACD. – Michael Dec 13 '17 at 7:54
• I hate to sound like everyone's math teacher, but this is definitely a case where making a reasonable scale drawing will guide your intuition for the formal proof. – jwg Dec 14 '17 at 11:33

Though Robert gave a great solution, I post another solution with a picture. Draw a line $\overline{DE}$ parallel to $\overline{AB}$, toward $\overline{BC}$. Then $\triangle DEC$ becomes an equilateral triangle and $\square ABED$ becomes a parallelogram, and in turn rhombus which makes $\triangle BEC$ be isosceles triangle. Thus we have $\angle BED = 74^\circ$,$\angle DEC=60^\circ$, and $\angle BCE = 23^\circ$. Therefore, we have $\angle DCE = 60^\circ - 23^\circ = 37^\circ$.

• Very nice solution, this one too. How did you draw this figure? – samjoe Dec 13 '17 at 12:07
• @samjoe Thanks! I drew it with goodnotes4 app in my ipad. – Matholic Dec 13 '17 at 17:06
• This is the best solution IMO. +1 – justhalf Dec 13 '17 at 18:27

There is a point $E$ along $DC$ such that $\triangle ABE$ is equilateral (why? See my P.S.). Therefore $$\hat{ADC}=\frac{1}{2}(180^\circ-(74^\circ-60^\circ))=83^\circ\quad\text{and}\quad \hat{BCD}=\frac{1}{2}(180^\circ-(166^\circ-60^\circ))=37^\circ.$$

Note that your picture is misleading because the angle at $D$ should be acute!

P.S. Go backwards. Start form the equilateral triangle $\triangle ABE$. Then draw externally the isosceles triangle $\triangle AED$ with an angle at $A$ of $14^\circ$, and the isosceles triangle $\triangle BEC$ with an angle at $B$ of $106^\circ$. Then $D$, $E$ and $C$ are on the same line because $$\hat{AED}+\hat{AEB}+\hat{BEC}=83^\circ+60^\circ+37^\circ=180^\circ.$$

• No. Not along $DC$. How can you say that? – Jaideep Khare Dec 13 '17 at 7:51
• @Robert I was wrong earlier. If angle ADC is less than 90 (which is the case here) then you can be correct. But how did you know that before hand? – samjoe Dec 13 '17 at 8:24
• I did think you went backwards. if that's what you did. A nice solution indeed! – samjoe Dec 13 '17 at 8:48
• @RobertZ :Would you please ,make details more clear ? Possible draw a figure ? (thanks again) – Khosrotash Dec 13 '17 at 10:01
• @Khosrotash I I did my best. Note that in your picture the angle at D should be acute! $\triangle ADE$ is isosceles!! – Robert Z Dec 13 '17 at 10:36

Only thing to realise here was that $\sin(\pi - x) = \sin (x)$. So $\sin DAC = \sin (\pi - DBC) = \sin DBC$. Also $AD = AB = BC$

Then we apply sine rule on $DAC$ and $DBC$ triangles.

$$\frac{\sin DCA}{ AD} = \frac{\sin DAC}{DC}$$ and $$\frac{\sin CDB}{ BC} = \frac{\sin DBC}{DC}$$

So we get angle $\angle DCA = \angle CDB = 30$ (seen from triangle $DOC$ where centre point is $O$) or the required angle as $37 ^\circ$.

This is somehow an ugly approach. Nevertheless, you can calculate $BD$ using the law of cosines. Then apply the law of cosines in $BDC$ to calculate $DC$ and use the law of sines to get the unknown angle.

• IMO using elementary geometry is not uglier than using trigonometry. – Rosie F Dec 13 '17 at 9:48
• @RosieF you may even get some interesting results, such as :)) $$\sin 37=\sin 67 \sqrt{\frac{1 - \sin 16}{3/2-\sin 16+\sqrt{2}\sin 23\sqrt{1-\sin 16}}}$$ – polfosol Dec 13 '17 at 10:42
• @Polfosol definitely +1 for that discovery :) – samjoe Dec 14 '17 at 13:11
• @polfosol:From where ? you can get this formula ...can you elaborate more please ? – Khosrotash Dec 15 '17 at 8:46
• @Khosrotash I am on mobile, so I should make it as brief as possible. If you follow the procedure I described in my answer, you will see that sin BCD / sin DBC = BD / CD and since DBC =67, BCD =37 we end up with that formula. – polfosol Dec 15 '17 at 18:33