Difference between revisions of "1996 AJHSME Problems/Problem 24"
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== Solution 2 == | == Solution 2 == | ||
Contruct <math>\overline{BE}</math> through <math>D</math> and intersects <math>\overline{AC}</math> at point <math>E</math> | Contruct <math>\overline{BE}</math> through <math>D</math> and intersects <math>\overline{AC}</math> at point <math>E</math> | ||
+ | |||
By Exterior Angle Theorem, | By Exterior Angle Theorem, | ||
− | <math>angle{ADE}</math> | + | |
+ | <math>\angle{ADE}</math> <math>=</math> <math>\angle{ABD} + \angle{BAD}</math> | ||
+ | |||
+ | Similarly, | ||
+ | |||
+ | <math>\angle{EDC} = \angle{DBC} + \angle{BCD}</math> | ||
+ | |||
+ | Thus, | ||
+ | |||
+ | <math>\angle{ADC} = \angle{ABC} + \angle{BAD} + \angle{BCD}</math> | ||
+ | |||
+ | Let <math>\angle{ADC} = x</math> | ||
+ | |||
+ | Because <math>\overline{AD}</math> and <math>\overline{CD}</math> are angle bisectors, | ||
+ | |||
+ | <math>180^\circ - x = \angle{BAD} + \angle{BCD}</math> | ||
+ | |||
+ | <math>= x - 50^\circ</math> | ||
+ | |||
+ | <math>x = 115^\circ</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 |
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