# Sine Law or No?

The diagram shows a triangle $ABC$ where $$AB = AC,\, BC = AD \text{ and } \angle BAC = 20°.$$ Find $\angle ADB$. I used the Sine Law; We know that $\sin(C)/\sin(BDC) = \sin(A)/\sin(ABD)$ If we let $\sin BDC = a$, then the equation will be equal to $\sin 80/\sin a = \sin20/\sin(a-20)$, I could not find any correlation with the the angle a and the other angles, is there a way to solve this using the sine law or is it a bad approach in general (or is the implementation of the sine law wrong?). What other approach would be a lot more useful in this kind of problem?

• Hmm.. I saw a way to get 4 linear equations with 4 variables but every time I try to solve them I just get circular dependencies for some reason. – tilper Nov 6 '17 at 13:59
• It's also worth mentioning that SIMOC is not currently active, so this isn't a contest problem that shouldn't be posted. Or, even better, this is from the 2015 SIMOC and has a final answer posted already anyway. – tilper Nov 6 '17 at 14:00
• Link to the contest: drive.google.com/file/d/0B0-8qjwNLbpgTGxJclpsNjhZUFU/view , btw the answer is 150 degrees ;Dj – SuperMage1 Nov 6 '17 at 14:20
• What are you actually trying to do? – Chase Ryan Taylor Nov 6 '17 at 14:45
• Oh I am Very sorry the edit messed up as it did not show the question (thanks for the edit though) But I am Finding Angle ADB, – SuperMage1 Nov 6 '17 at 14:52

## 3 Answers

Ok, let start with your equation $$\frac{\sin 80}{\sin a}=\frac{\sin 20}{\sin(a-20)}.$$ Using twice equation $\sin(2x)=2\sin x\cos x$, we reach $$\frac{4\sin 20\cos 20\cos 40 }{\sin a}=\frac{\sin 20}{\sin(a-20)}$$ or $$\frac{4\cos 20\cos 40}{\sin a}=\frac{1}{\sin(a-20)}.$$ We know that $2\cos20\cos40=\cos60+\cos20$, thus $$\frac{1+2\cos 20}{\sin a}=\frac{1}{\sin(a-20)}.$$ Simplifying, $$\sin a=\sin(a-20)+2\cos20\sin(a-20)=\sin(a-20)+\sin a+\sin(a-40)$$ which turns to $$\sin(a-20)+\sin(a-40)=0.$$ This equation has $a=30$ as a solution, thus $\sphericalangle BDA=150$.

• Is there a solution where in it does not involve the sum-product identity, just the basic trig? Or do we really have to use it in order to answer this question? – SuperMage1 Nov 6 '17 at 14:51
• I really have no idea. In solving this question, I tried to use cosine law which was almost work but I didn't go on. – Khayyam Nov 6 '17 at 15:58

You can solve it without any trig at all, but it requires some "inspired" additional constructions. Reflect the triangle across one of its sides, $AC$, and also let $BL$ be the angle bisector of $\angle ABC$: All the labelled angles in the diagram are easy to calculate using properties of isosceles triangles and/or sums of angles in triangles, using the reflection symmetry and/or the bisector. For example, $\angle ABK = 70^\circ$ because $\Delta ABK$ is isosceles with vertex $\angle BAK = 40^\circ$.

Now consider $\Delta KLM$ and $\Delta KLC$ - they are congruent by two angles and a side ($KL$ which they have in common), therefore $KM = KC = BC = AD$ (the blue segments; the second equality holds because of the reflection). Also, now we can see $BM = AM = CD$ (the first from the isosceles $\Delta ABM$; the second, from $AD + DC = AC = AB = AK = AM + MK$).

Finally, see that $\Delta BCD$ and $\Delta KMB$ have two sides equal (the blue and red segments) and the $80^\circ$ angle between them as well, so they are congruent, too. From this, $\angle BDC = \angle KBM = 30^\circ$.

Alternative construction and demonstration with simple Euclidean geometry below. Sorry for the label change. Here we have $$AC \cong BC$$, $$CP \cong AB$$ and $$\angle ACB = 20°$$.

• Draw two isosceles triangles congruent to $$ABC$$ and with one side in common, so that $$\triangle ABC \cong \triangle BCD \cong \triangle DCE$$.

• Show that $$\triangle ACE$$ is equilateral.

• Determine $$\angle EAB = \angle CAB - \angle CAE$$.
• Demonstrate that $$\triangle CPB \cong \triangle EAB$$, with SAS criterion. Therefore $$\angle CPB \cong \angle ABE$$.
• Show that $$\triangle DBE$$ is isosceles.
• Thus calculate the measure of the angle $$\angle DBE = \frac{180°-80°}{2}$$.
• So finally $$\angle CPB = \angle ABE = \angle ABD - \angle ABE = 150°$$.

An even simpler construction is reported below ($$\triangle ADB$$ is equilater; show then that $$\triangle ADC \cong \triangle CDB \cong \triangle CPB$$ and deduce from there the result). 