# 5: IIT LEVEL SOLVED II

What you need to differentiate here is the difference between an INTERCEPT and PERPENDICULAR DISTANCE

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

This is an interesting question very naturally constructed out of rotation of a point about another point.

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.

This is an important skill to have and IIT Entrance level questions often tests the students on their grasp over this concept.

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)