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>180^\circ - x = \angle{BAD} + \angle{BCD}</math> |
| − | + | <math>= x - 50^\circ \\$ | |
| − | + | </math>x = 115^\circ<math> | |
| − | + | Thus, the answer is </math>\boxed{C}$ | |
| − | |||
| − | |||
| − | |||
| − | Thus, the answer is <math>\boxed{C} | ||
~ lovelearning999 | ~ lovelearning999 | ||
Revision as of 07:40, 6 October 2024
Contents
Problem
The measure of angle 
is 
, 
bisects angle 
, and 
bisects angle 
. The measure of angle 
is
![[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,N); label("$50^\circ $",(9.4,8.8),SW); [/asy]](http://latex.artofproblemsolving.com/b/8/c/b8c508ab617894b8e5319d685f1b7a1a01defa24.png)
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,
$= x - 50^\circ \$ (Error compiling LaTeX. Unknown error_msg)x = 115^\circ\boxed{C}$
\boxed{C}$
~ 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. 
