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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
3 years 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, 12 months ago

Thanks

- 2 years, 12 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, 12 months ago

From sine law in triangle ADC we have AD/2R=sin30°=1/2,so AD=R. That is hiding behind his result.That is why we locate circum-centre of triangle ADC.Circumcentre of tringle ADC is outside of the triangle,A and O are on the oposite sides of line CD.
When we apply sine law in triangle ADC to line CD we have that CD/2R=sin110°,so CD=2Rsin110°=2Rsin70°=2Rcos20° ,but AB=CD,so AB=2Rcos20°,from sine law aplied to the triangle ABD(which has circumcentre r) and line AD we get AD=2rsinB, so R=2rsinB,so sinB=R/(2r).But when apply sin law to the triangle ABC and line AB we have AB=2rsin140°=2rsin40° so 2rsin40°=2Rcos20°,and we get R/r=sin40°/cos20°.Now we replace it and get sinB=0,5(sin40°/cos20)=0,5(2sin20°cos20°/cos20°)=sin20°,so b=20° or b=160°,but B<90° so answer is B=20°.

- 2 years, 10 months ago