Difference between revisions of "2017 AMC 8 Problems/Problem 6"
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~MathFun1000 | ~MathFun1000 | ||
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Since we see the ratio is <math>3:3:4</math>, we can rule out the answer of <math>{\textbf{(E) }90}</math> because the numbers in the ratio are too big to have <math>90^\circ</math>. Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is <math>\boxed{\textbf{(D) }72}</math>. | Since we see the ratio is <math>3:3:4</math>, we can rule out the answer of <math>{\textbf{(E) }90}</math> because the numbers in the ratio are too big to have <math>90^\circ</math>. Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is <math>\boxed{\textbf{(D) }72}</math>. | ||
Revision as of 10:23, 17 February 2023
Contents
Problem
If the degree measures of the angles of a triangle are in the ratio 
, what is the degree measure of the largest angle of the triangle?
Solution 1
The sum of the ratios is 
. Since the sum of the angles of a triangle is 
, the ratio can be scaled up to 
The numbers in the ratio represent the angles of the triangle. The question asks for the largest, so the answer is 
.
Solution 2
We can denote the angles of the triangle as 
, , 
. Due to the sum of the angles in a triangle, 
. The greatest angle is and after substitution we get .
~MathFun1000
Solution 4 (Brute Force) NOT RECOMMENDED
Since we see the ratio is , we can rule out the answer of 
because the numbers in the ratio are too big to have 
. Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is .
~jason.ca
Video Solution
https://youtu.be/rQUwNC0gqdg?t=635
~savannahsolver
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
