Write a tangent ratio for each angle of elevation in terms of the height $h$, $x$, and $x+12$.

For complementary acute angles, the tangent of one angle is the reciprocal of the tangent of the other. Use this to relate $h^2$ to the two horizontal distances.

Apply the Pythagorean theorem to the right triangle that starts at $Q$ and has hypotenuse $4\sqrt{35}$.

First use the complementary angles to derive $h^2=x(x+12)$. Algebra is the fastest way to establish this relationship.

Enter $y=x(x+12)+(x+12)^2$ and $y=560$. Select the intersection with positive $x$ to find the distance from $R$ to the base.

Using the positive intersection value $x=8$, enter $\sqrt{8(8+12)}$ to calculate the antenna height.

Let $x$ be the distance from $R$ to $B$. Since $Q$, $R$, and $B$ are in that order and $QR=12$, the distance from $Q$ to $B$ is $x+12$.

Let $h$ be the antenna height. The tangent of the angle at $Q$ is $\frac{h}{x+12}$, and the tangent of the angle at $R$ is $\frac{h}{x}$. Because the angles are complementary, their tangents are reciprocals. Therefore, $\frac{h}{x}=\frac{x+12}{h}$, so $h^2=x(x+12)$.

The triangle from $Q$ to the top of the antenna has legs $h$ and $x+12$ and hypotenuse $4\sqrt{35}$. Thus, $h^2+(x+12)^2=560$. Substitute $h^2=x(x+12)$ to get $x(x+12)+(x+12)^2=560$.

Let $y=x+12$. Then $x=y-12$, so $(y-12)y+y^2=560$. This simplifies to $y^2-6y-280=0$, giving $y=20$ because a distance must be positive. Thus $x=8$, and $h^2=8(20)=160$. Therefore, $h=4\sqrt{10}$.