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>. 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> (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> |
==See Also== | ==See Also== |
Revision as of 00:55, 5 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
(3\cdot 18:3\cdot 18:4\cdot 18). 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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