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# HELP!

Does anyone have a non-trigonometry solution for this problem:

Suppose we have a triangle $$ABC$$.It is known that $$C=30$$ degrees.From $$A$$ draw a segment $$AD$$ such that $$D$$ is on $$BC$$ and $$ADC$$ is $$40$$ degrees.Also $$AB=CD$$.Find $$B$$.

Note by Lawrence Bush
2 years, 7 months ago

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My thinking is as follows: Locate the circum-centre of triangle ADC . Let's call it O. /AOD=2*/ACD=2*30 or 60°. This makes triangle AOD equilateral or AD=CO = R, the circumradius of Tr. ADC. Now in triangles BAD & DCO, we've AB=CD (Given), AD=CO & /_ BDA=/DOC both =140° . This isn't the included angle but since it is clearly the largest angle of each triangle, this is a special case of congruence. This means /ABD or /_ B = /_CDO = 20° · 2 years, 7 months ago

Thanks · 2 years, 7 months ago

Amazing! You're awesome @One Top How did you think of it? I though of drawing the circumcircle of $$\Delta ABC$$... · 2 years, 7 months ago