1990 AHSME Problems/Problem 3

Problem

The consecutive angles of a trapezoid form an arithmetic sequence. If the smallest angle is $75^\circ$, then the largest angle is

$\text{(A) } 95^\circ\quad \text{(B) } 100^\circ\quad \text{(C) } 105^\circ\quad \text{(D) } 110^\circ\quad \text{(E) } 115^\circ$

Solution

A trapezoid is a quadrilateral; therefore the interior angles sum to $360^\circ$. Thus $75+(75+x)+(75+2x)+(75+3x)=360$, so $x=10$ and the largest angle is $105^\circ$ which is $\fbox{C}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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