2015 AMC 12B Problems/Problem 13

Revision as of 16:06, 6 March 2015 by Pi over two (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$. What is $AC$?

$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$

Solution

$\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$, hence $\angle ACB = \angle ADB = 40^\circ$. By considering $\triangle ABC$, it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$. Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{\textbf{(B)}\; 6}.$

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS