1998 AMC 12 Problem 26

Attempt Problem 26 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

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26.

In quadrilateral ABCD,ABCD, it is given that ∠A=120∘,\angle A=120^\circ, angles BB and DD are right angles, AB=13,AB=13, and AD=46.AD=46. Then AC=AC=
6060
6262
6464
6565
7272
Answer: B
Concepts:cyclic quadrilateralslaw of cosinescircumradius
Small hint:
The two right angles make ABCDABCD cyclic with ACAC as a diameter
Big hint:
Find BDBD in triangle ABDABD, then use the extended law of sines
Solution:
Because ∠B+∠D=180∘,\angle B+\angle D=180^\circ, the quadrilateral is cyclic, and ACAC is its diameter. In triangle ABD,ABD, BD2=132+462−2(13)(46)cos⁡120∘=2883=(313)2. \begin{aligned} BD^2&=13^2+46^2\\ &\quad-2(13)(46)\cos120^\circ\\ &=2883=(31\sqrt3)^2. \end{aligned} By the extended law of sines, AC=BDsin⁡120∘AC=\frac{BD}{\sin120^\circ} =31332=62.=\frac{31\sqrt3}{\frac{\sqrt3}{2}}=62. Thus B is correct.
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