Calculate Torque About Axis B: F, θ, ϕ, π Involved

[PerpetuallyConfused]

Homework Statement

What is the torque τB about axis B due to the force F⃗ ? (B is the point at Cartesian coordinates (0,b), located a distance b from the origin along the y axis.)
Express the torque about axis B in terms of F, θ, ϕ, π, and/or other given coordinate data.

Homework Equations

[/B]
τ = RF sin (θ)

The Attempt at a Solution

So I know that the answer is Tb = bFsin (pi/2+θ)
What I don't understand is how to get pi/2+θ?
I know that θ is the angle between r vector and F vector and since that's the case shouldn't the answer be pi/2-θ?

 

[Orodruin]

Can you please show us your own effort? What did you get?

 

[PerpetuallyConfused]

Can you please show us your own effort? What did you get?

Well I thought the answer was Tb = bFsin (pi/2-θ) but that's obviously wrong. I would like to know why. My homework says that alpha is the angle between r vector and F vector therefore I thought that it would be pi/2-θ.

 

[Orodruin]

The angle $\alpha$ is drawn in the figure. Note that the vector $\vec r$ is the vector from the point relative to which you want to know the torque to the point where the force acts. In this case the vector from B to A.

Edit: Also note that, since what matters is the component of $\vec F$ perpendicular to $\vec r$, it does not matter if you use $\alpha$ or $\pi - \alpha$ since
$$
\sin(\pi - \alpha) = \sin(\pi)\cos(\alpha) - \cos(\pi)\sin(\alpha) = 0 + \sin(\alpha) = \sin(\alpha).
$$
It holds that
$$
\sin(\pi/2 - \theta) = \sin(\pi/2)\cos(\theta) - \sin(\theta)\cos(\pi/2) = \cos(\theta)\sin(\pi/2)
= \cos(\theta)\sin(\pi/2) + \sin(\theta)\cos(\pi/2) = \sin(\pi/2 + \theta).
$$