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 $B$.

Easy Math Editor

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Comments

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°

Ajit Athle - 6 years, 7 months ago

Amazing! You're awesome @One Top How did you think of it? I though of drawing the circumcircle of $\Delta ABC$ ...

Satvik Golechha - 6 years, 7 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°.

Nikola Djuric - 6 years, 6 months ago

Thanks

lawrence Bush - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$