# 2: Rotate Square & Eq T

A square is rotated about the Origin by x degrees in the clockwise direction. Can you find the coordinates of the vertices now.

What we need to do is to see the angle x and find other angles which are also x and then locate right angled triangles containing x and finally use trigonometry to find the sides of the right angled triangles and thus the coordinates of the vertices of the square.

So, basically
1. find angles equal to x
2. find right angled triangles with x
3. use aCosx and aSinx

Remember that the corners of the square are 90 degrees.

We have taken the straight line x/a+y/b=1 and in the right angled triangle so formed with the axes, we draw three squares and are to find the coordinates of the vertices of the square. Like the previous example, the procedure is the same.
We have taken the sides of the large squares as p which is acutally equal to a^2+b^2

The important fact here is the height of an Equilateral triangle: it is basically aSin60 and this you have to remember always because often it is used.

In the second diagram, we have two lines of the triangle for which we have found the equations.

We have used a trigonometric formula that says

Sin (180-x)=Sin x
Cos (180-x)= – Cos x
tan (180-x)= – tan x

and we have used the theorem for finding the equation of a line passing through (p,q) with slope, m is y – q = m (x – p)

Here we have taken the same equilateral triangle and put it inside a circle. This is a very useful way to reckon any point. Later you will study the parametric form of the equation of a straight line which uses the concept of the circle.

We then turn the equilateral triangle by angle x in the clockwise direction like we had done for the square in an earlier example.

The equilateral triangle has now been drawn on the intercept of the line x/a+y/b=1 between the axes. The length of the hypotenuse of the right angled triangle so formed has been named p.

There is a mistake in the diagram. Can you spot it?

Hint: it is in the coordinates of V