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Exercise 247
The angles of quadrilateral $ABCD$ satisfy $\angle A=2\angle B=3\angle C=5\angle D$. What is the degree measure of $\angle A$, rounded to the nearest whole number?
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Solution
Let $\angle A=2\angle B=3\angle C=5\angle D=k$, so $\angle B=\frac k2$, $\angle C=\frac k3$, $\angle D=\frac k5$.
$k\left(1+\frac12+\frac13+\frac15\right)=\frac{61}{30}k=360^\circ$, so $k=\dfrac{10800}{61}\approx177.05^\circ$.
Rounded, $\angle A\approx177^\circ$.
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