Difference between revisions of "1996 AJHSME Problems/Problem 24"
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Because <math>\overline{AD}</math> and <math>\overline{CD}</math> are angle bisectors, | Because <math>\overline{AD}</math> and <math>\overline{CD}</math> are angle bisectors, | ||
− | <math>\ | + | <math>180^\circ - x = \angle{BAD} + \angle{BCD}</math> |
− | <math>= x - \ | + | <math>= x - 50^\circ</math> |
− | <math>x = \ | + | <math>x = 115^\circ</math> |
==See Also== | ==See Also== |
Revision as of 20:33, 1 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,
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 |
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