Difference between revisions of "1996 AJHSME Problems/Problem 24"
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(There we go) |
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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, | ||
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | 180^\circ - x &= \angle{BAD} + \angle{BCD}\\ | ||
+ | &=x - 50^\circ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
− | + | <math>x = 115^\circ</math> | |
− | + | Thus, the answer is <math>\boxed{C}</math> | |
− | |||
− | Thus, the answer is < | ||
~ lovelearning999 | ~ lovelearning999 |
Latest revision as of 14:27, 4 August 2025
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.