Q1: P is a point inside a square ABCD of side s, and |PA| = 5, |PB| = 3, and |PC| = 7. Find s.
Q2: P is a point inside a hexagon ABCDEF of side s, and |PA| = 2, |PF| = 3, and |PE|=4. Find s.
Can somebody please explain how to do these with maths, and also explain in English how they are done? Thanks!
Well I know about pythagorus and sin cos tan kind of, but I dont understand thm really. And I dont know about the laws of cosines. What I dont understand about your post is how you knew that you had to rotate the triangle about point P. And how you knew that you had to use the law of cosines and how you know that angle PBE is 45 degrees.
I think you mean that angle EPB is 45 degrees. This you can know as tan EPB is 1(side EB is three and side BP is also three, and that they're meeting at a 90 degree angle, which gives tan EPB= EB/BP= 3/3=1), and when tan is 1, the angle is automatically 1 (if it's not larger than 180 degrees). In order to understand the rest of the explanation given above by Archie Meade you need to learn about trigonometry in arbitrary triangles and about the law of cosines.
I moved the triangle ABP to it's new position, to allow solution
using the Law of Cosines.
Non-right-angled triangles are typically solved using the Law of Sines or The Law of Cosines.
If you are given a side and it's opposite angle and a third dimension (or clue to obtaining a third)
of a non-right-angled triangle,
then the Law of Sines can easily find all other dimensions.
If you have not got a side length and the measure of it's opposite angle
then we may resort to the Law of Cosines
if we can find 3 dimensions.
This is why I moved the triangle to it's new position
as I could easily find the third side of triangle BPE,
giving me all 3 sides of triangle PCE.
This allowed me to then calculate the angle "theta",
which gives me 3 dimensions of triangle BPC,
hence any other measurement of that triangle can then be found,
in particular the length of the side BC.
Rotate triangle AFP about the point F until A rests on E and P on G.
Triangle FPG is isosceles and angle GFP is
This is because the side FP rotates through the same angle as FA.
Hence angles GPF and PGF are both
Next find the length of the line GP using the Law of Cosines
This allows to be found using triangle PGE...
Rearrange the above Law of Cosines equation to solve for
having substituted in the value of
Finally, using the Law of Cosines for triangle EFP