Difference between revisions of "2011 AMC 10B Problems/Problem 17"
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− | We could also tell that quadrilateral <math>BEDC</math> | + | We could also tell that quadrilateral <math>BEDC</math> is an isosceles trapezoid because for <math>\overline{BE}</math> and <math>\overline{DC}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 12:43, 27 July 2018
Contents
[hide]Problem
In the given circle, the diameter is parallel to , and is parallel to . The angles and are in the ratio . What is the degree measure of angle ?
Solution 1
We can let be and be because they are in the ratio . When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem, and .
because they are alternate interior angles and . Opposite angles in a cyclic quadrilateral are supplementary, so . Use substitution to get
Note:
We could also tell that quadrilateral is an isosceles trapezoid because for and
Solution 2
Note as before. The sum of the interior angles for quadrilateral is . Denote the center of the circle as . . Denote and . We wish to find . Our equation is . Our final equation becomes . After subtracting and dividing by , our answer becomes
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
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