Difference between revisions of "2009 AMC 10B Problems/Problem 9"
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Finally, <math>\triangle CDE</math> is isosceles, hence each of the two remaining angles (<math>\angle D</math> and <math>\angle E</math>) is equal to <math>\frac{180^\circ - 75^\circ}2 = \frac{105^\circ}2 = \boxed{52.5}</math>. | Finally, <math>\triangle CDE</math> is isosceles, hence each of the two remaining angles (<math>\angle D</math> and <math>\angle E</math>) is equal to <math>\frac{180^\circ - 75^\circ}2 = \frac{105^\circ}2 = \boxed{52.5}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/hsP804ZSocg | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 18:46, 18 January 2021
Contents
Problem
Segment and intersect at , as shown, , and . What is the degree measure of ?
Solution
is isosceles, hence .
The sum of internal angles of can now be expressed as , hence , and each of the other two angles is .
Now we know that .
Finally, is isosceles, hence each of the two remaining angles ( and ) is equal to .
Video Solution
~savannahsolver
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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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