Difference between revisions of "2017 AMC 8 Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | The sum of the ratios is <math>10</math>. Since the sum of the angles of a triangle is <math>180^{\circ}</math>, the ratio can be scaled up to <math>54:54:72</math> (3\cdot 18:3\cdot 18:4\cdot 18). The numbers in the ratio <math>54:54:72</math> represent the angles of the triangle. We want the largest, so the answer is <math>\boxed{\textbf{(D) }72}</math> | + | The sum of the ratios is <math>10</math>. Since the sum of the angles of a triangle is <math>180^{\circ}</math>, the ratio can be scaled up to <math>54:54:72</math> <math>(3\cdot 18:3\cdot 18:4\cdot 18)</math>. The numbers in the ratio <math>54:54:72</math> represent the angles of the triangle. We want the largest, so the answer is <math>\boxed{\textbf{(D) }72}</math> |
==See Also== | ==See Also== | ||
Revision as of 23:55, 4 January 2018
Problem 6
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
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. We want the largest, so the answer is 
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 | ||
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