Answer
The force exerted on the person's back muscles is $$F\approx408.1lb$$
Work Step by Step
$$F=\frac{0.6W\sin(\theta+90^\circ)}{\sin12^\circ}$$
We can expand $\sin(\theta+90^\circ)$ using the sine sum identity, which states
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
So, $$\sin(\theta+90^\circ)=\sin\theta\cos90^\circ+\cos\theta\sin90^\circ$$
$$\sin(\theta+90^\circ)=\sin\theta\times0+\cos\theta\times1$$
$$\sin(\theta+90^\circ)=\cos\theta$$
Therefore, $F$ can be rewritten as $$F=\frac{0.6W\cos\theta}{\sin12^\circ}$$
$$W=200lb\hspace{3cm}\theta=45^\circ$$
Apply the given information to the formula of $F$:
$$F=\frac{0.6\times200lb\times\cos45^\circ}{\sin12^\circ}$$
$$F=\frac{120\times\cos45^\circ}{\sin12^\circ}$$
Now we use calculator for calculation.
$$F\approx408.1lb$$
