The altitudes of a triangle are $12, 15,$ and $20.$ The largest angle in this triangle is $\text{(A) }72^\circ \qquad$ $ \text{(B) }75^\circ \qquad$ $ \text{(C) }90^\circ \qquad$ $ \text{(D) }108^\circ \qquad$ $ \text{(E) }120^\circ$
Solution
If $a,b,c$ are the sides opposite altitudes $12,15,20$ and $K$ is twice the area, then $a=\frac{K}{12}, b=\frac{K}{15}, c=\frac{K}{20}$, so $a:b:c=\frac1{12}:\frac1{15}:\frac1{20}=5:4:3$.
Since $3^2+4^2=5^2$, the triangle is (similar to) a $3$-$4$-$5$ right triangle, with the right angle opposite the side of ratio $5$.
