Difference between revisions of "1982 AHSME Problems/Problem 6"

Revision as of 12:27, 16 July 2024

Problem

The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$ . The remaining angle is

$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)}\ 130^\circ\qquad \text{(E)}\ 144^\circ$

Solution

Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\circ$ , and each interior angle belongs to $[0, 180^\circ)$ . Therefore, we must have $n - 2 = \lfloor \frac{2570^\circ}{180^\circ} \rfloor = 15$ . Then the missing angle must be $180*15^\circ - 2570^\circ = 130^\circ$ , so our answer is $\boxed{\text{(D)}}$ and we are done.

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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		+	== Problem ==
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	The sum of all but one of the interior angles of a convex polygon equals <math>2570^\circ</math>. The remaining angle is		The sum of all but one of the interior angles of a convex polygon equals <math>2570^\circ</math>. The remaining angle is

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Difference between revisions of "1982 AHSME Problems/Problem 6"

Revision as of 12:27, 16 July 2024

Problem

Solution

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