Problem:
In quadrilateral ABCD, it is given that β A=120β, angles B and D are right angles, AB=13, and AD=46. Then AC=
Answer Choices:
A. 60
B. 62
C. 64
D. 65
E. 72
Solution:
Extend DA through A and CB through B and denote the intersection by E. Triangle ABE is a 30ββ60ββ90β triangle with AB=13, so AE=26. Triangle CDE is also a 30ββ60ββ90β triangle, from which it follows that CD=(46+26)/3β=243β. Now apply the Pythagorean Theorem to triangle CDA to find that AC=462+(243β)2β=62.
OR
Since the opposite angles sum to a straight angle, the quadrilateral is cyclic, and AC is the diameter of the circumscribed circle. Thus AC is the diameter of the circumcircle of triangle ABD. By the Extended Law of Sines,
AC=sin120βBDβ=3β/2BDβ.
We determine BD by the Law of Cosines:
BD2=132+462+2β 13β 46β 21β=2883=3β 312, so BD=313β.