# Some hard questions. Please explain how to go about doing them.

• Feb 13th 2011, 08:58 AM
JoshuaJava
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!
• Feb 13th 2011, 11:04 AM
Quote:

Originally Posted by JoshuaJava
Q1: P is a point inside a square ABCD of side s, and |PA| = 5, |PB| = 3, and |PC| = 7. Find s.

Can somebody please explain how to do these with maths, and also explain in English how they are done? Thanks!

Q1

Rotate triangle APB about B until it is triangle BCE.

Triangle BPE is a right-angled isosceles triangle, hence

$\displaystyle |PE|^2=3^2+3^2$

We could use the Law of Cosines if we had the angle $\displaystyle \theta$

$\displaystyle s^2=|BC|^2=3^2+7^2-2(3)7cos\left(\theta+45^o\right)$

We can find $\displaystyle \theta$ from triangle PCE...

$\displaystyle 5^2=7^2+|PE|^2-(2)7|PE|cos\theta$

Solve for $\displaystyle \theta$

Finally solve for s.
• Feb 13th 2011, 11:32 AM
JoshuaJava
Thanks for the help, but I still don't know what you're talking about.
• Feb 13th 2011, 11:58 AM
JoshuaJava
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.
• Feb 13th 2011, 12:32 PM
scounged
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.
• Feb 13th 2011, 01:16 PM
Quote:

Originally Posted by JoshuaJava
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.

You need to understand them or you will have difficulty.
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.
• Feb 13th 2011, 02:52 PM
JoshuaJava
Thanks a lot! This really helped!
• Feb 14th 2011, 03:05 AM
Quote:

Originally Posted by JoshuaJava

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!

This time, the angle between sides of the regular hexagon is $\displaystyle 120^o$

Rotate triangle AFP about the point F until A rests on E and P on G.

Triangle FPG is isosceles and angle GFP is $\displaystyle 120^o$

This is because the side FP rotates through the same angle as FA.

Hence angles GPF and PGF are both $\displaystyle 30^o$

Next find the length of the line GP using the Law of Cosines

$\displaystyle |GP|^2=3^2+3^2-2(3)3cos120^o$

This allows $\displaystyle \theta$ to be found using triangle PGE...

$\displaystyle 2^2=|PG|^2+4^2-2|PG|4cos\theta$

Rearrange the above Law of Cosines equation to solve for $\displaystyle \theta$
having substituted in the value of $\displaystyle |PG|$

Finally, using the Law of Cosines for triangle EFP

$\displaystyle s^2=|FE|^2=3^2+4^2-2(3)4cos\left(30^o+\theta\right)$
• Feb 16th 2011, 08:05 AM
Wilmer
Quote:

Originally Posted by JoshuaJava
Thanks a lot! This really helped!

So, for practice, try this one:
point P is inside rectangle ABCD: AP=52, BP=25, CP=39 and DP=60.
Find the side lengths of rectangle ABCD.