SOLUTION: Given that {{{tan(theta)=3/5}}} find all possible coordinates for point P(x,y) in the unit circle

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Question 1084769: Given that tan%28theta%29=3%2F5 find all possible coordinates for point P(x,y) in the unit circle
Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+theta+ can be in the 1st or 3rd quadrants
where the tan function is positive.
+c%5E2+=+a%5E2+%2B+b%5E2+
+c%5E2+=+3%5E2+%2B+5%5E2+
+c%5E2+=+9+%2B+25+
+c%5E2+=+34+
+c+=+sqrt%2834%29+
----------------------
On the unit circle,
+c+=+1+
+a+=+3%2Fsqrt%2834%29+
+b+=+5%2Fsqrt%2834%29+
-----------------------
In the 1st quadrant, the point is:
P( 3/sqrt(34), 5/sqrt(34) )
In the 3rd quadrant, the point is:
P( -3/sqrt(34), -5/sqrt(34) )
-------------------------------
get a 2nd opinion on this if needed

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!

Here's a different approach for finding the two points P(x,y)

Since tangent=opposite%2Fadjacent, draw a right triangle whose
opposite side is the numerator of 3/5, which is 3, and whose adjacent 
side is the denominator of 3/5, which is 5.  Then θ will be the 
angle with opposite side 3 and adjacent side 5:



Calculate the hypotenuse by the Pythagorean theorem:

c%5E2=a%5E2%2Bb%5E2
c%5E2=5%5E2%2B3%5E2
c%5E2=25%2B9
c%5E2=34
c=sqrt%2834%29



We place the triangle on a graph so that the vertex is at the origin,
and draw a circle with center at the origin.  But this circle is not
the unit circle.



The unit circle has radius 1, but the above circle has radius √34, so we
divide everything by √34, to make the circle become the unit circle,
so we have the first value of P(x,y):



Also, since tangent is positive in QIII, we can also reflect everything
across the origin, and the angle is increased by 180°, and get another
point P(x,y).



Edwin