# In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$. Where's my error?

In $$\triangle ABC$$ with $$AB=AC$$ and $$\angle BAC=20^\circ$$, point $$D$$ is on $$AC$$, with $$BC=AD$$. Find $$\angle DBC$$.

I know the correct solution, but I'm more interested in where is the problem in my solution.

My solution :

Now in $$\triangle ABD$$, applying the sine rule:

$$\frac{AD}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{1}$$

In $$\triangle BDC$$:

$$\frac{BD}{\sin 80^\circ} = \frac{BC}{\sin(180^\circ-\beta)} \tag{2}$$

We know $$AD= BC$$; put in to $$(1)$$:

$$\frac{BC}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{3}$$

Comparing $$(2)$$ and $$(3)$$:

$$\frac{BC}{BD} = \frac{\sin\alpha}{\sin 20^\circ} = \frac{\sin(180^\circ-\beta)}{\sin 80^\circ} \tag{4}$$

$$\frac{\sin \alpha}{\sin(180^\circ-\beta)} = \frac{\sin 20^\circ}{\sin 80^\circ} \tag{5}$$

Now, $$\alpha = 20^\circ$$ and $$\beta = 100^\circ$$, but when I plug these values in $$\triangle ABC$$, it's not even triangle. oO

Where I am wrong? Thanks.

PS : sorry for poor editing, I don't have any clue about it.

• Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers. – астон вілла олоф мэллбэрг Jan 19 at 9:27
• @AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :) – Blue Jan 19 at 9:31
• Good to see you taking the message on board. As for the actual question, from $\frac{\sin(\alpha)}{\sin(180-\beta)} = \frac{\sin 20}{\sin 80}$, how did you get to $\alpha = 20$ and $\beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $\alpha = 20$ and $180-\beta = 80$ so $\beta = 100$. That must be the incorrect step here, then. – астон вілла олоф мэллбэрг Jan 19 at 9:31
• I see. But that does not work here, unfortunately. Instead, note that $\sin (180 - \beta) = \sin \beta$ from the sum of sines formula, and also that $\beta + \alpha = 160$ so $160 - \alpha = \beta$. So substituting, gets you $\frac{\sin \alpha}{\sin(160-\alpha)} = \frac{\sin 20}{\sin 80}$. See what you can do from here. Also, since you know the correct value of $\alpha$, check that it satisfies the above equation. – астон вілла олоф мэллбэрг Jan 19 at 9:37
• @Blue Thanks for helping with editing. Much appreciation sir :) – Angelus Mortis Jan 19 at 10:59

## 1 Answer

So we have that $$\frac{\sin \alpha}{\sin (180-\beta)} = \frac{\sin 20}{\sin 80}$$.

The first thing we use is that $$\alpha + \beta = 160$$ from the triangle $$ABD$$. From here, $$180 - \beta = 180 - (160-\alpha) = 20 + \alpha$$.

Next, we note that: $$\frac{\sin 20}{\sin 80} = \frac{\sin 20}{\cos(90-80)} = \frac{\sin 20}{\cos 10} = \frac{2 \sin 10 \cos 10} {\cos 10} = 2 \sin 10$$

So, we have the equation : $$\frac{\sin \alpha}{\sin (\alpha + 20)} = 2 \sin 10\\ \implies \sin \alpha = 2 \sin 10 \sin (20+\alpha) = 2 \sin 10 \sin 20 \cos \alpha + 2 \sin 10 \cos 20 \sin \alpha$$

Now, collecting terms of $$\sin \alpha$$ on one side, and facctorizing it out, $$\sin \alpha(1 - 2 \sin 10 \cos 20) = 2 \sin 10 \sin 20 \cos \alpha \\ \implies \tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20}$$

The right hand side is some fixed number which we have to find.

To do this, we first simplify the denominator, using the formulas : $$2 \sin A\cos B = \sin(A+B) + \sin(A-B) \quad ; \quad \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

We will also use the fact that $$\sin 30 = \frac 12$$. In our case, $$1 - 2 \sin 10 \cos 20 = 1- (\sin 30 + \sin (-10)) = 2 \sin 30 - (\sin 30 - \sin 10) \\ = \sin 30 +\sin 10 = 2 \sin 20 \cos 10$$

Therefore, $$\tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20} = \frac{2\sin 10 \sin 20}{2 \cos 10 \sin 20} = \tan 10$$

Now, since $$0 < \alpha < 180$$, we get that $$\alpha = 10$$. From here, $$80-\alpha = 70$$ is the desired angle.