Difference between revisions of "1993 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | |
+ | We first consider <math>\angle CBA</math>. Because <math>\angle A = 55</math> and <math>\angle C = 75</math>, <math>\angle B = 180 - 55 - 75 = 50</math>. Then, because <math>\triangle BED</math> is isosceles, we have the equation <math>2 \angle BED + 50 = 180</math>. Solving this equation gives us <math>\angle BED = 65 \rightarrow \fbox{D}</math> | ||
== See also == | == See also == |
Revision as of 08:24, 9 August 2018
Problem
In , , is on side and is on side . If , then
Solution
We first consider . Because and , . Then, because is isosceles, we have the equation . Solving this equation gives us
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AHSME Problems and Solutions |
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