Difference between revisions of "1996 AJHSME Problems/Problem 24"
(→Solution 2) |
(→Solution 2) |
||
(5 intermediate revisions by the same user not shown) | |||
Line 53: | Line 53: | ||
Thus, the answer is <math>\boxed{C}</math> | Thus, the answer is <math>\boxed{C}</math> | ||
+ | |||
+ | ~ lovelearning999 | ||
==See Also== | ==See Also== |
Latest revision as of 07:41, 6 October 2024
Contents
Problem
The measure of angle is , bisects angle , and bisects angle . The measure of angle is
Solution
Let , and let
From , we know that , leading to .
From , we know that . Plugging in , we get , which is answer .
Solution 2
Contruct through and intersects at point
By Exterior Angle Theorem,
Similarly,
Thus,
Let
Because and are angle bisectors,
Thus, the answer is
~ lovelearning999
See Also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.