1982 AHSME Problems/Problem 6

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 $(n-2) \cdot 180^\circ$ , and each interior angle is less than $180^\circ$ . Therefore, $n - 2 = \left\lfloor \frac{2570^\circ}{180^\circ} \right\rfloor = 15$ , and the missing angle is $15 \cdot 180^\circ - 2570^\circ = 130^\circ \boxed{\text{(D)}}$ .

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1982 AHSME Problems/Problem 6

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