Difference between revisions of "1962 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | {{ | + | Calculating for angles <math>\angle DBC</math> and <math>\angle DCB</math>, we get |
+ | |||
+ | <math>\angle DBC</math> = <math>90 - \frac{B}{2}</math> and | ||
+ | <math>\angle DCB</math> = <math>90 - \frac{C}{2}</math>. | ||
+ | |||
+ | In triangle BCD, we have | ||
+ | |||
+ | <math>\angle BDC</math> = <math>180 - (90 - \frac{B}{2}) - (90 - \frac{C}{2})</math> = <math>\frac{B+C}{2}</math> = <math>\frac{1}{2}\cdot(180 - A)</math>, so the answer is <math>\fbox{C}</math>. | ||
==See Also== | ==See Also== |
Revision as of 23:32, 28 August 2018
Problem
Let the bisectors of the exterior angles at and of triangle meet at D Then, if all measurements are in degrees, angle equals:
Solution
Calculating for angles and , we get
= and = .
In triangle BCD, we have
= = = , so the answer is .
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AHSME Problems and Solutions |
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