SOLUTION: Give a point (5,-5)on the terminal side of angle 0 find the value of tan.

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: Give a point (5,-5)on the terminal side of angle 0 find the value of tan.      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 207936: Give a point (5,-5)on the terminal side of angle 0 find the value of tan.
Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
Given a point (5,-5)on the terminal side of angle q find the value of tanq

Plot the point (5,-5)

drawing%28400%2C400%2C-7%2C7%2C-7%2C7%2C%0D%0A%0D%0Agraph%28400%2C400%2C-7%2C7%2C-7%2C7%29%2C+%0D%0A%0D%0Aline%285%2B.1%2C-5%2C5-.1%2C-5%29%2Cline%285%2C-5%2B.1%2C5%2C-5-.1%29%2Cline%285%2B.1%2C-5-.1%2C5-.1%2C-5%2B.1%29%2Cline%285-.1%2C-5%2B.1%2C5%2B.1%2C-5-.1%29%2C+locate%285%2C-5%2C%27%285%2C-5%29%27%29%29

Draw the radius vector from the origin to
the point (5,-5):

drawing%28400%2C400%2C-7%2C7%2C-7%2C7%2C%0D%0Aline%285-.3%2C-5%2C5%2C-5%29%2Cline%285%2C-5%2C5%2C-5%2B.3%29%2C%0D%0Agraph%28400%2C400%2C-7%2C7%2C-7%2C7%29%2C+%0D%0Aline%280%2C0%2C5%2C-5%29%2C+locate%285%2C-5%2C%27%285%2C-5%29%27%29%29

Indicate with a curved line the angle q measured 
counter-clockwise from the right side of the 
x-axis around to the radius vector.

drawing%28400%2C400%2C-7%2C7%2C-7%2C7%2C%0D%0Aline%285-.3%2C-5%2C5%2C-5%29%2Cline%285%2C-5%2C5%2C-5%2B.3%29%2C%0D%0Agraph%28400%2C400%2C-7%2C7%2C-7%2C7%29%2C+%0D%0Aarc%280%2C0%2C2.5%2C-2.5%2C0%2C315%29%2C%0D%0Aline%280%2C0%2C5%2C-5%29%2C+locate%285%2C-5%2C%27%285%2C-5%29%27%29%29

Draw the x and the y coordinates, forming
a right triangle:

drawing%28400%2C400%2C-7%2C7%2C-7%2C7%2C+locate%282%2C.6%2C%27x=5%27%29%2C+locate%285.1%2C-2.1%2C%27y=-5%27%29%2C%0D%0Aline%285-.3%2C-5%2C5%2C-5%29%2Cline%285%2C-5%2C5%2C-5%2B.3%29%2C%0D%0Agraph%28400%2C400%2C-7%2C7%2C-7%2C7%29%2C+%0D%0Aarc%280%2C0%2C2.5%2C-2.5%2C0%2C315%29%2C%0D%0Aline%280%2C0%2C5%2C-5%29%2C+locate%285%2C-5%2C%27%285%2C-5%29%27%29%2C%0D%0Atriangle%280%2C0%2C5%2C-5%2C5%2C0%29+%29

Draw another arc (the red one below) to
represent the REFERENCE angle, the one
with vertex at the origin which is
inside the triangle:

drawing%28400%2C400%2C-7%2C7%2C-7%2C7%2C%0D%0Aline%285-.3%2C-5%2C5%2C-5%29%2Cline%285%2C-5%2C5%2C-5%2B.3%29%2Clocate%282%2C.6%2C%27x=5%27%29%2C+locate%285.1%2C-2.1%2C%27y=-5%27%29%2C%0D%0Agraph%28400%2C400%2C-7%2C7%2C-7%2C7%29%2C+%0D%0Aarc%280%2C0%2C2.5%2C-2.5%2C0%2C315%29%2C%0D%0Aline%280%2C0%2C5%2C-5%29%2C+locate%285%2C-5%2C%27%285%2C-5%29%27%29%2C%0D%0Atriangle%280%2C0%2C5%2C-5%2C5%2C0%29%2C%0D%0Ared%28arc%280%2C0%2C3.6%2C-3.6%2C315%2C360%29%29%0D%0A%0D%0A+%29

Since 

TANGENT=%28OPPOSITE%29%2F%28ADJACENT%29

The side OPPOSITE the reference angle (red arc) is 
y=-5 and the ADJACENT side to the reference
angle is x=5, then the tangent of the angle in
standard position which is q,
(black arc), has the tangent y%2Fx, so we 
end up with

tanq = y%2Fx=%28-5%29%2F5=-1

Edwin