Torque about a pendulum's suspension point

[JessicaHelena]

Homework Statement

In the figure attached, what is the torque about the pendulum's suspension point produced by the weight of the bob, given that the mass is 40 cm to the right of the suspension point, measured horizontally, and m=0.50kg?

Homework Equations

tau = rFsin (theta)
or
tau = lF

The Attempt at a Solution

I don't know why but I tend to prefer the first equation, so upon seeing this, I thought the F_g = mg is pulling down on the suspension point at an angle of 60°, so I can do tau = L*m*g*sin(60°).
From the given information about the 40cm, we can draw a right triangle and get that L = 0.80m.
sin 60° is sqrt(3)/2.
m = 0.5 kg
g is a constant 9.8.
Then multipling them all together, I got 1.7Nm.

However, apparently I am wrong — it should be 2.0 Nm, and no, it isn't a matter of rounding 1.7Nm up to 2.0Nm. The solution provides a way of using the second equation, but I'd like to know why I was wrong using the 1st equation. Could someone help me out?

 

[BvU]

I don't know why but I tend to prefer the first equation

Good (or perhaps not good...) ! But what is $\theta$ ? and F ?

all variables and given/known data

A picture showing F and $\theta$ would really help

 

[haruspex]

an angle of 60°, so I can do tau = L*m*g*sin(60°).

Think that through again.