1998 AHSME Problem 26

Geometry Level 2

In quadrilateral ABCD, it is given that A = 12 0 \angle A = 120^{\circ} , angles B and D are right angles, AB = 13, and AD = 46. Then AC=

62 60 65 64

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1 solution

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Since B = C = 9 0 , \angle B = \angle C = 90^{\circ}, A C AC is a diameter of A B C D . \circledcirc ABCD. By cosine rule on A B D , \triangle ABD, we find out that B D = A B 2 + A D 2 2 A B A D cos A = 2883 . BD= \sqrt{AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos \angle A} = \sqrt{2883}. By Sine rule on A B D , \triangle ABD, B D sin A = A C A C = 2 2883 3 = 62 . \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

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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

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