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Two congruent circles, centered at $A$ and $B$, intersect at $C$. When $AC$ is extended, it intersects the circle centered at $B$ at $D$. If $\angle DAB$ is $43^{\circ}$, then find $\angle DBA$, in degrees.

What I attempted to do was to project a line from $A$ upward until it touched the circle and connect it with $C$. I then extend a line from $C$ to $B$. I though these two new triangles were congruent due to similar sides and found the $\theta$ above $A$ to be $47^{\circ}$ so thought $D$ would also be $47^{\circ}$ leaving the final $\angle DBA$ to be $90^{\circ}$ which is not correct.

Thanks for any help my geometry is quite bad which is why I'm trying to do problems and look up relevant information when I get stuck. I don't have much information so I know I have to extend some lines but other then that I'm unsure.

Jack D'Aurizio
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HighSchool15
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    The reason that your attempt didn't work is that the point at the top of the circle doesn't actually lie on the line through $B$ and $C$. So when you connected that point to $C$, then drew a line from $C$ to $B$, you would have had a quadrilateral, not a triangle. – Théophile Aug 08 '15 at 04:17

2 Answers2

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Both $CAB$ and $CBD$ are isosceles triangles, so: $$\widehat{ADB}=\widehat{DCB}=2\cdot\widehat{CAB}=86^\circ $$ and: $$\widehat{DBA}=\widehat{CBA}+\widehat{CBD}=43^\circ+(180^\circ-2\cdot 86^\circ) = 43^\circ+8^\circ = \color{red}{51^\circ}.$$

Jack D'Aurizio
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$\color{red}{\text{Solution by using trigonometry}}$

Let, $R$ be radius of each circle & $\angle DBA=x\implies \angle ADB=180^\circ-(43^\circ+x)$

In isosceles $\triangle BCD$ $$CD=2R\cos (180^\circ-(43^\circ+x))=-2R\cos(43^\circ+x)$$ $$AD=AC+CD=R-2R\cos (43^\circ+x)=R(1-2\cos(43^\circ+x))$$

Now, applying sine rule in $\triangle ABD$ $$\frac{\sin \angle DAB}{BD}=\frac{\sin \angle DBA}{AD}$$ Now, substituting the corresponding values we get $$\frac{\sin 43^\circ}{R}=\frac{\sin x}{R(1-2\cos(43^\circ+x))}$$

$$\sin 43^\circ(1-2\cos (43^\circ+x))=\sin x$$

$$\sin 43^\circ-2\sin 43^\circ\cos (43^\circ+x)=\sin x$$

$$(1-2\sin^243^\circ)\sin x+\sin 86^\circ\cos x=\sin 43^\circ$$

$$\sin x\cos 86^\circ+\sin 86^\circ\cos x=\sin 43^\circ$$ $$\sin (x+86^\circ)=\sin 43^\circ$$ But, $x>0$ hence we get $$x+86^\circ=180^\circ-43^\circ$$ $$x=180^\circ-43^\circ-86^\circ=51^\circ$$ Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\angle DBA=51^\circ}}$$

Harish Chandra Rajpoot
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