Difference between revisions of "1954 AHSME Problems/Problem 37"
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<math>p+ q + 2\theta = 180</math> | <math>p+ q + 2\theta = 180</math> | ||
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<math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180</math> | <math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180</math> | ||
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<math>p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}</math> | <math>p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}</math> | ||
Revision as of 17:23, 15 April 2017
Problem 37
Given with bisecting , extended to and a right angle, then:
Solution
Let be .
Partial Solution
Looking at triangle PRQ, we have and from the given statement , so looking at triangle MOR , which rules out
1954 AHSC (Problems • Answer Key • Resources) | ||
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Followed by Problem 39 | |
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