Difference between revisions of "2000 AMC 8 Problems/Problem 13"
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− | In triangle <math>CAT</math>, we have <math> \angle ACT =\angle ATC </math> and <math> \angle CAT = 36^\circ </math>. If <math> \overline{TR} </math> bisects \angle ATC <math>, then < | + | ==Problem== |
+ | |||
+ | In triangle <math>CAT</math>, we have <math> \angle ACT =\angle ATC </math> and <math> \angle CAT = 36^\circ </math>. If <math> \overline{TR} </math> bisects <math>\angle ATC </math>, then <math> \angle CRT = </math> | ||
<asy> | <asy> | ||
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<math> \text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ </math> | <math> \text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | In <math>\triangle ACT</math>, the three angles sum to <math>180^\circ</math>, and <math>\angle C = \angle T</math> | ||
+ | |||
+ | <math>\angle CAT + \angle ATC + \angle ACT = 180</math> | ||
+ | |||
+ | <math>36 + \angle ATC + \angle ATC = 180</math> | ||
+ | |||
+ | <math>2 \angle ATC = 144</math> | ||
+ | |||
+ | <math>\angle ATC = 72</math> | ||
+ | |||
+ | Since <math>\angle ATC</math> is bisected by <math>\overline{TR}</math>, <math>\angle RTC = \frac{72}{2} = 36</math> | ||
+ | |||
+ | Now focusing on the smaller <math>\triangle RTC</math>, the sum of the angles in that triangle is <math>180^\circ</math>, so: | ||
+ | |||
+ | <math>\angle RTC + \angle TCR + \angle CRT = 180</math> | ||
+ | |||
+ | <math>36 + \angle ACT + \angle CRT = 180</math> | ||
+ | |||
+ | <math>36 + \angle ATC + \angle CRT = 180</math> | ||
+ | |||
+ | <math>36 + 72 + \angle CRT = 180</math> | ||
+ | |||
+ | <math>\angle CRT = 72^\circ</math>, giving the answer <math>\boxed{C}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=12|num-a=14}} |
Revision as of 20:11, 30 July 2011
Problem
In triangle , we have
and
. If
bisects
, then
Solution
In , the three angles sum to
, and
Since is bisected by
,
Now focusing on the smaller , the sum of the angles in that triangle is
, so:
, giving the answer
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |