AMC10 2007 10B, Problem 15

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?

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} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \fbox{173}\]

Answer: 173