Difference between revisions of "2011 AMC 10B Problems/Problem 17"
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== Solution == | == Solution == | ||
− | We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by | + | We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by triangle sum theorem, <math>4x+5x+90=180 \longrightarrow x=10</math> and <math>\angle ABE = 50</math>. |
− | <math>\angle ABE = \angle BED</math> because they are | + | <math>\angle ABE = \angle BED</math> because they are alternate interior angles and <math>\overline{AB} \parallel \overline{ED}</math>. Opposite angles in a [[cyclic]] quadrilateral are [[supplementary]], so <math>\angle BED + \angle BCD = 180</math>. Use substitution to get <math>\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}</math> |
== See Also== | == See Also== |
Revision as of 12:38, 10 September 2017
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
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
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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