Geometry Problem - 3

Geometry Level pending

Triangle A B C ABC is isosceles with A B = A C AB=AC . Given that E B C = 4 0 , E B D = 3 0 , D C E = 4 0 \angle EBC=40^\circ, \angle EBD=30^\circ, \angle DCE=40^\circ and D C B = 3 0 \angle DCB=30^\circ , find D E B \angle DEB correct to four decimal places.


The answer is 24.3737.

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1 solution

Since B E C = 180 40 30 40 = 70 \angle BEC=180-40-30-40=70 , E B C \triangle EBC is isosceles with B C = B E BC=BE .

C D B = 180 30 30 40 = 8 0 \angle CDB=180-30-30-40=80^\circ

Let B C = B E = 1 BC=BE=1

By sine law on E B C \triangle EBC , we have

C E sin 40 = B C sin 70 \dfrac{CE}{\sin 40}=\dfrac{BC}{\sin 70}

C E = sin 40 sin 70 CE=\dfrac{\sin 40}{\sin 70}

By sine law on D B C \triangle DBC , we have

D C sin 70 = B C sin 80 \dfrac{DC}{\sin 70}=\dfrac{BC}{\sin 80}

D C = sin 70 sin 80 DC=\dfrac{\sin 70}{\sin 80}

B D sin 30 = B C sin 80 \dfrac{BD}{\sin 30}=\dfrac{BC}{\sin 80}

B D = sin 30 sin 80 BD=\dfrac{\sin 30}{\sin 80}

By cosine law on E D C \triangle EDC , we have

D E 2 = ( sin 40 sin 70 ) 2 + ( sin 70 sin 80 ) 2 2 ( sin 40 sin 70 ) ( sin 70 sin 80 ) ( cos 40 ) DE^2=\left(\dfrac{\sin 40}{\sin 70}\right)^2+\left(\dfrac{\sin 70}{\sin 80}\right)^2-2\left(\dfrac{\sin 40}{\sin 70}\right)\left(\dfrac{\sin 70}{\sin 80}\right)(\cos 40)

D E 0.615132148 DE\approx 0.615132148

By sine law on D E B \triangle DEB , we have

sin D E B sin 30 sin 80 = sin 30 0.615132148 \dfrac{\sin \angle DEB}{\dfrac{\sin 30}{\sin 80}}=\dfrac{\sin 30}{0.615132148}

D E B 24.3737 \angle DEB \approx \color{#D61F06}\boxed{24.3737}

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