Difference between revisions of "2015 AMC 12B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | + | <math>\angle ADB</math> and <math>\angle ACB</math> are both subtended by segment <math>AB</math>, hence <math>\angle ACB = \angle ADB = 40^\circ</math>. By considering <math>\triangle ABC</math>, it follows that <math>\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ</math>. Hence <math>\triangle ABC</math> is isosceles, and <math>AC = BC = \boxed{\textbf{(B)}\; 6}.</math> | |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=14|num-b=12}} | {{AMC12 box|year=2015|ab=B|num-a=14|num-b=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:06, 6 March 2015
Problem
Quadrilateral 
is inscribed in a circle with 
and 
. What is 
?
Solution
and 
are both subtended by segment 
, hence 
. By considering 
, it follows that 
. Hence is isosceles, and 
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 | |
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