2007 AMC 10B Problems/Problem 15

Problem

The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?

$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$

Solution

The sum of the interior angles of any quadrilateral is $360^\circ.$ $\begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*}$ $\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{\mathrm{(D) \ } 173}$

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2007 AMC 10B Problems/Problem 15

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