further we have used a slightly advanced formula for tan (A+B) and that the parallel lines both have the slope as – 2 and since we know the perpendicular distance through the parallel line distance formula, we can immediately locate a right angled triangle to find Sin Cos tan of the angle between the lines as shown
As the slope of PQ is found to be 1 implies PQ makes 45 degree with the x direction.
And when PQ is turned about P by 60 it makes 45+60 degrees. And if you subtract 90 degrees you end up with 75 degrees.
Knowing that tan 15 is [2 – sqrt 3], we have computed Sin 15 & Cos 15 and used them to find our solution, which looks a little involved.
If you closely follow the example shown, it reveals the entire method. Additionally you need to know the formulae for Cos (A+B) & Sin (A+B)