Answer
$$45 d h \sqrt{\left(\frac{b-a}{2}\right)^{2}+h^{2}}$$
Work Step by Step
Since
$$
\sin (\theta)=\frac{\text { Side Opposite }}{\text { Hypotenuse }}=\frac{h}{\sqrt{\left(\frac{b-a}{2}\right)^{2}+h^{2}}}
$$
Therefore,
\begin{aligned}
F=w \int_{0}^{h} \frac{d \cdot y}{\sin (\theta)} d y &=\frac{90 d}{\sin (\theta)} \int_{0}^{h} y d y \\
&=\frac{90 d}{\sin (\theta)}\left[\frac{y^{2}}{2}\right]_{0}^{h} \\
&=\frac{45 d h^{2}}{\sin (\theta)} \\
&=45 d h \sqrt{\left(\frac{b-a}{2}\right)^{2}+h^{2}}
\end{aligned}
