
vector angle
two forces $\displaystyle F_1$N and $\displaystyle F_2$ N are given by
$\displaystyle F_1 = 24i + 32j  42k$
$\displaystyle F_2 = 3i  72j  30k $
a) calculate, correct to the nearest degree, the angle between the directions of $\displaystyle F_1$ and $\displaystyle F_2$
so using the dot product formula $\displaystyle F_1 \cdot F_2 = F_1F_2cos\theta$
I end up with $\displaystyle \frac{1332}{58 * 3\surd677} = cos\theta$
$\displaystyle \therefore \theta = 72.89 $
I think i have done this right however the angle i inverse cos is negative, so do I draw up a cast diagram to find the angle? if so is it the angle 180  72.98?

I think that you made a mistake in the dot product. I get $\displaystyle 1116$.
Recall that if the dot product is negative the angle is obtuse.

yes i miscalculated. i now get the angle 75.7.
so I'm thinking I do 18075.7 = 104.3 ?