Difference between revisions of "2015 AMC 12B Problems/Problem 13"
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==Problem== | ==Problem== | ||
| + | Quadrilateral <math>ABCD</math> is inscribed in a circle with <math>\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,</math> and <math>BC=6</math>. What is <math>AC</math>? | ||
| − | + | <math>\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</math> | |
==Solution== | ==Solution== | ||
Revision as of 00:49, 4 March 2015
Problem
Quadrilateral 
is inscribed in a circle with 
and 
. What is 
?
Solution
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
