Difference between revisions of "1982 AHSME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | Note that the sum of the interior angles of a convex polygon of <math>n</math> sides is <math> | + | Note that the sum of the interior angles of a convex polygon of <math>n</math> sides is <math>(n-2) \cdot 180^\circ</math>, and each interior angle is less than <math>180^\circ</math>. Therefore, <math>n - 2 = \left\lfloor \frac{2570^\circ}{180^\circ} \right\rfloor = 15</math>, and the missing angle is <math>15 \cdot 180^\circ - 2570^\circ = 130^\circ \boxed{\text{(D)}}</math> . |
== See also == | == See also == | ||
Latest revision as of 21:15, 29 June 2025
Problem
The sum of all but one of the interior angles of a convex polygon equals 
. The remaining angle is
Solution
Note that the sum of the interior angles of a convex polygon of 
sides is 
, and each interior angle is less than 
. Therefore, 
, and the missing angle is 
.
See also
| 1982 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
