1998 AHSME Problem 26

Geometry Level 2

In quadrilateral ABCD, it is given that $\angle A = 120^{\circ}$ , angles B and D are right angles, AB = 13, and AD = 46. Then AC=

62 60 65 64

1 solution

Sreejato Bhattacharya
Mar 28, 2014

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Since $\angle B = \angle C = 90^{\circ},$ $AC$ is a diameter of $\circledcirc ABCD.$ By cosine rule on $\triangle ABD,$ we find out that $BD= \sqrt{AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos \angle A} = \sqrt{2883}.$ By Sine rule on $\triangle ABD,$ $\dfrac{BD}{\sin \angle A} = AC \implies AC = \dfrac{2 \sqrt{2883}}{\sqrt{3}} = \boxed{62}.$

But sir!! angle B and D are right angles You took angle B and C right angles so how is that correct?

Kandarp Kakkad - 6 years ago

It follows directly from the "Extended Sine Law" (see wiki) which relates a triangle to the radius of a circumscribed circle.

I just learned about that now. I solved it from first principles and sum angle formula.

Matt O - 5 years, 6 months ago

1998 AHSME Problem 26

1 solution

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