How do I find DB in this triangle?
1 Answer
You can use law of sines. Note that ADC=105 and hence ACD=45. Using law of sines in the triangle ADC:
\[\frac{CD}{\sin 30}=\frac{AD}{\sin 45} \Rightarrow CD=AD \frac{\sin 30}{\sin 45}=2\frac{\sin 30}{\sin 45}\]
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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