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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 :

Geo figure

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.

Blue
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Angelus Mortis
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    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. – Sarvesh Ravichandran Iyer Jan 19 '19 at 09:27
  • @астонвіллаолофмэллбэрг will surely do Sir :) – Angelus Mortis Jan 19 '19 at 09:29
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    @AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :) – Blue Jan 19 '19 at 09:31
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    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. – Sarvesh Ravichandran Iyer Jan 19 '19 at 09:31
  • @Blue duly noted sir :) – Angelus Mortis Jan 19 '19 at 09:32
  • @астонвіллаолофмэллбэрг It's how I have been taught to compare values in sine rule. If that step is wrong , then , how to proceed towards answer with aplha and beta ? – Angelus Mortis Jan 19 '19 at 09:34
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    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. – Sarvesh Ravichandran Iyer Jan 19 '19 at 09:37
  • Related: "Find an angle of an isosceles triangle". The question there is identical to the one here, but since OP is looking for an error in a particular solution, it's not a duplicate. – Blue Jan 19 '19 at 09:40
  • @астонвіллаолофмэллбэрг then doesn't aplha get two values, 20 and 80 ? , so we need to plug values check? – Angelus Mortis Jan 19 '19 at 09:41
  • It seems that none of these values work. What is the correct answer to this question, since you know it? Let us try to work backwards, and check which equation it satisfies from the ones you have written above. – Sarvesh Ravichandran Iyer Jan 19 '19 at 09:48
  • Indeed, $\alpha = 10$ fits the bill, giving $\beta = 150$. This satisfies even the last equation. What this means is that while you have not made mistakes, we have landed up with an equation involves the sine of two different angles, hence is quite difficult to solve. – Sarvesh Ravichandran Iyer Jan 19 '19 at 09:58
  • @астонвіллаолофмэллбэрг Indeed α = 10 in the solution. But I'm still baffled :( – Angelus Mortis Jan 19 '19 at 10:58
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    @Blue Thanks for helping with editing. Much appreciation sir :) – Angelus Mortis Jan 19 '19 at 10:59
  • @AngelusMortis I have managed to continue from where you left off, and solved the trigonometric equation. Can I write the solution? – Sarvesh Ravichandran Iyer Jan 19 '19 at 13:00
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    @астонвіллаолофмэллбэрг Sir please do. I'm waiting eagerly :) – Angelus Mortis Jan 20 '19 at 04:51
  • If you mean angle $BDC$ by $(180-\beta)$,it is not correct it is $BDC=[180^o-(80 +\beta)]=(100-\beta)$. – sirous Jan 20 '19 at 15:05

1 Answers1

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

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.

Sarvesh Ravichandran Iyer
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