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Trigonometry
By Matthew Wood (P3881) on Friday, February 16, 2001 - 07:35 pm:

HELP
I am stuck with some coursework.

Can I work out all 3 sides of a triangle when I only have ALL 3 angles

What will the Area to Perimeter Ratio be if the angles are 40°, 60° and 80°?

By Pooya Farshim (P2572) on Friday, February 16, 2001 - 09:17 pm:

The answer to the first question is no.
Consider a triangle of angles 45, 45, 90 degrees.
If you draw a square of any given side and then draw a diameter then the triangles formed satisfy the conditions given in the problem but there are many of them as you can choose the side of the square to be any number.

2nd question:
Let the sides be a,b,c, and then try and find b and c in terms of a.
One formula for the area of a triangle says that area=½absinC. Using this,
A = ½×a×b×sin(40)
A = ½×a×c×sin(80)
A = ½×b×c×sin(60)

Equating (2) and (3):
b=a×(sin 80/sin 60)
Equating (1) and (3):
c=a×(sin 40/sin 60)

Therefore:

Area = ½×a×a×(sin 80/sin 60)×sin(40)
Perimeter = a + a×(sin 80/sin 60) + a×(sin 40/sin 60)
so
A/P= ½×a×(sin 80/sin 60)×sin(40)/(1+(sin 80/sin 60)+(sin 40/sin 60))
(cancelling a in top and bottom)
So:
A/P=½×a×sin(80)×sin(40)/(sin(60)+sin(40)+sin(80))

(a is the side between the angles 80 and 40)

By Matthew Wood (P3881) on Saturday, February 17, 2001 - 10:36 am:

My Maths teacher said something about letting 1 side be k, and working the others out in terms of k
eg k1.375, because the ratio between lines will always stay the same. Is it possible to work out the area to Perimeter Ratio, as a number, by this method.

By Pooya Farshim (P2572) on Saturday, February 17, 2001 - 12:40 pm:

Hi Matthew:

That's exactly what I have done except that I have used "a" instead of "k".

To clarify it a bit more:

Let the sides of the triangle be a, b and c. (we will find b and c in terms of a in a moment). The formulae for the area of the triangle is: half times two sides times the sin of the angle between them. So let's say the angle between the sides a and b is 40, a and c is 80 and b and c is 60. Let's write the formulae for area:

A=½×a×b×sin(40)
A=½×a×c×sin(80)
A=½×b×c×sin(60)

dividing (2) by (3) gives:
1=(a/b)×(sin 80/sin 60) => b=a× (sin 80/sin 60)
So we have find b in terms of a. (sin 80/sin 60) is just a number (roughly 1.137)

dividing (1) by (3):
1=(a/c)×(sin 40/sin 60) => c=a×(sin 40/sin 60)

So we have find c in terms of a as well. ((sin 40/sin 60) is roughlly 0.742)

Let's put back the value of the b in the the first formuale for area:

A=½×a×b×sin(40)
A=½×a×a×(sin 80/sin 60)×sin(40)
But perimeter=a+b+c
Let's put the values of b and c in this formulae as well:
P=a+a×(sin 80/sin 60)+a×(sin 40/sin 60)

Now A/P=(½×a×a×(sin 80/sin 60)×sin(40))/(a+a×(sin 80/sin 60)+a×(sin 40/sin 60))

cancelling a's give:

A/P=½×a×sin(80)×sin(40)/(sin(60)+sin(40)+sin(80))

All the stuff involving sin(??) are just some numbers and you can find them using your calculator (mine gives roughly .1269) (don't forget to multiply the ½ as well)

Therefore: ratio of area to Perimeter=a×.1269)

By Matthew Wood (P3881) on Saturday, February 17, 2001 - 12:58 pm:

Thank you so much, Pooya