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 sides of the large squares as p which is acutally equal to a^2+b^2

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)

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

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

Hint: it is in the coordinates of V

See the answer below

We have also used advanced formulae for Sin (A+B) & Cos (A+B)

The x coordinate of V will be [a + pCosy], not pCosy as shown.