How do you find the sides of a triangle when only angles are given
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How do you find the sides of a triangle when only angles are given

[From: ] [author: ] [Date: 12-06-04] [Hit: ]
google.com/images…However, this is not what they are asking.They are simply asking for coordinate of a point so that the angle from with positive x-axis and line from origin to the point gives the angle.I would guess this is what you have:http://staff.jccc.......
The Question is:
For each angle below, determine the exact coordinates of a point on the terminal arm of the angle in standard position.

a) 30*
b)45*
c)60*

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First, you cannot find the sides of a triangle when only angles are given.
You can have 2 similar triangles with the same angles, but with different side lengths:
https://encrypted-tbn0.google.com/images…

However, this is not what they are asking.
They are simply asking for coordinate of a point so that the angle from with positive x-axis and line from origin to the point gives the angle.

I would guess this is what you have:
http://staff.jccc.edu/rgrondahl/Unit_Cir…

This is fairly easy to find:

x = r cosθ
y = r sinθ

where r = radius of circle, θ = angle with positive x-axis

However, r is not provided, so I assume the points lie on unit circle, i.e. r = 1

a)

x = cos(30) = √3/2
y = sin(30) = 1/2

b)

x = cos(45) = 1/√2
y = sin(45) = 1/√2

c)

x = cos(60) = 1/2
y = sin(60) = √3/2

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those angles dont make a triangle, the measure of all of the angles has to add up to 180 degrees, since those dont the triangle you would draw would not form a triangle at all. also without being given any of the lengths of a triangle side it is impossible to find any of the side lengths, all that is possible is to find the ratios of the side lengths with respects to one specific side, if this indeed were a real triangle and on of the side was known from their you could use law of sines to find the other lengths which is "SINa/A =SINb/B =SINc/C" where lower case letters are the angles and upper case letters are the sides. if that doesn't work it is necessary to use law of cosines which is "A^2=B^2+C^2-2(B)(C)COSa" where again upper case is the side lengths and lower case is the angle.
EDIT: now its a 30, 60, 90 triangle: the trick for these is that they are all the same,the side opposite of the 30 angle will be x in length, the hypotenuse will be 2X in length, and the other remain sied will be X multiplied by the square root of three
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