# math

Quad ABCD with diagonal DB. <adc=90, <c=91, <cbd=43. segment ad=segment dc=9.
Which segment is longer ab or ad and why?

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1. by the law of sines, the side opposite the larger angle has the larger length.

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posted by Steve
2. How are you solving for <abd. Making assumptions?

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posted by Jacob
3. by law of sines,

bd/sin91 = 9/sin43
bd = 13.19

by law of cosines,

ab^2 = 9^2 + bd^2 - 2(9)(bd)cos44
= 81 + 174 - 170.84
ab = 9.17

Looks like ab > ad

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posted by Steve
4. Thank you steve, but something must be wrong with this problem. We have not learned law of cosines or law of sines yet and the correct answer is ad?? Baffled at how to get this based on what we are studying, SAS and SSS inequality theorems.

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posted by Jacob
5. Hmmm. If <C were also 90, then DA would be parallel to BC. In that case triangles ADB and DCB would be congruent, and ABCD would be a square, with ad=ab. But that would also require <cbd to be 45, not 43.

So, since <cbd is less than 45, bc > ad, so also ab > ad.

I still don't get ad > ab.

Also, since SAS and SSS are for congruency, not sure how they apply in determining inequality here.

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posted by Steve

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