# How do I find DB in this triangle?

## 1 Answer

\[\frac{CD}{\sin 30}=\frac{AD}{\sin 45} \Rightarrow CD=AD \frac{\sin 30}{\sin 45}=2\frac{\sin 30}{\sin 45}\]

Using law of sines in the triangle BDC:

\[\frac{CD}{\sin 45}=\frac{BD}{\sin 60} \Rightarrow CD=BD \frac{\sin 45}{\sin 60}.\]

Hence

\[CD=2\frac{\sin 30}{\sin 45}=BD \frac{\sin 45}{\sin 60}\]

\[\Rightarrow BD=2\frac{\sin 30 \sin 60}{(\sin45)^2}=2\frac{\frac{1}{2}\frac{\sqrt{3}}{2}}{(\frac{\sqrt{2}}{2})^2}=\sqrt{3}.\]

Erdos

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