# How to find the work done by a tilted force in a block as is pulled upwards in an incline?

The problem is as follows:

The figure from below shows a block which is pulled from point $$A$$ to point $$B$$ by a force $$F=50\,N$$ and a constant direction. Find the work in Joules that is made between points $$A$$ and $$B$$. (Hint: You may use the triangle $$7-24-25$$ for $$16^{\circ}-74^{\circ}-90^{\circ}$$)

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&-310\,J\\ 2.&-250\,J\\ 3.&+310\,J\\ 4.&+250\,J\\ 5.&+280\,J\\ \end{array}$$

Since the angle they use is $$8^{\circ}$$.

I could use the identity for half angle to obtain the relationships in the given triangle.

$$\sin 8^{\circ}=\sqrt{\frac{1-\cos 16^{\circ}}{2}}=\sqrt{\frac{1-\frac{24}{25}}{2}}$$

$$\sin 8^{\circ}=\sqrt{\frac{\frac{1}{25}}{2}}=\frac{\sqrt{2}}{10}$$

$$\cos 8^{\circ}=\sqrt{\frac{\frac{49}{25}}{2}}=\frac{7\sqrt{2}}{10}$$

But other than that I'm still stuck.

I can tell the distance between $$A$$ and $$B$$ as:

$$AB=7\sec 8^{\circ}=\frac{7}{\cos 8^{\circ}}=\frac{7}{\frac{7\sqrt{2}}{10}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$$

However that's how far I went with this problem.

I'm stuck as I don't know how to use the information provided of the force with the angle given.

My intuition tells me that I could naively say okay:

$$W= F\times d = 50\cos 37^{\circ} \times 5 \sqrt{2}$$

But I'm certain that this will not be the answer and neither appears in the alternatives. Can somebody help me here please?.

angle $$\theta +53^o + 8^o = 90^o$$ by the Figure $$\theta = 29^o$$
So $$W = F \times d = 50 \cos 29^o \times 5 \sqrt{2} = 309.22476293882954$$