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Powerprep : Co-oridnate Geom, Triangles

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by simplyjat » Sat Apr 05, 2008 9:11 am
The question is discussed previously many times; search and you will get the explanation
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by lagomez » Sat Apr 05, 2008 9:38 am
what i did is get the radius 1st...

coordinates (- sqrt 3, and 1) are given..i formed a triangle with x = sqrt 3 as the base, 1 as the height...Pythagorean theorem gives hypotenuse of 2

(sqrt 3)^2 + 1^2 = sqrt 4 = 2

so the radius is 2

a half circle means the arc is 180 degrees

form an imaginary triangle with radii 2..that triangle has a measure of 90 degrees at the (0,0)......so the other two triangles must = 90 degrees

look at the smaller triangle you formed with base sqrt 3 and height 1..the formula that stands out is 1-sqrt 3-2 which is a 30-60-90 triangle...



if the height is 1 then the angle opposite one is the shorter angle or 30 degrees

so..from the large imaginary triangle of 90 degrees and the small one of 30 degrees..the imaginary triangle in quadrant 1 must be 60 degrees...since the radius is 2...and we know one angle is 60 and one is 90..the other must be 30 or equation 1 - sqrt 3 - 2 again..
since we know the arc is 60 degrees...

since 60 degrees is the angle of the arc the height must be sqrt 3 (or the height) and the 90 degrees is the radius or 2..the shortest leg (in this case the x-coordinate) is 1 (the 30 degree line)
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by lagomez » Sat Apr 05, 2008 9:42 am
if you get confused..let me know from where and i'll take it from there
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by mmukher » Sat Apr 05, 2008 10:36 am
Thanks, that helped.


The diagram as drawn made the figure look symmetrical wrt to the Y axis and that through me off.

The key was to guess given the sides as 1:sqrt3:2 as a 30-60-90 triangle on the left of the Y-axis.


Thanks, this was driving me nuts.
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