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.
$\endgroup$ 161 Answer
$\begingroup$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.
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