Grade 11 math
posted by Priya .
If the given point is on the terminal arm of an angle A in stadard position. 0 degrees more than or equal too A and less than or equal too 180 degree, find sin A and cos A:
a) (6,8) b) (-5,12)
Can you tell me how you get it because I do not understand it at all please help!
based on a right-angled triangle , you probably originally learned the following definitions of trig functions:
sinØ = opposite/hypotenuse
cosØ = adjacent/hypotenuse
tanØ = opposite/adjacent
by now I hope you also learned them in the following terms
sinØ = y/r
cosØ = x/r
tanØ = y/x
So after drawing your triangle for the terminal point (6,8)
we know x=6, y=8 and r^2 = x^2 + y^2 = 100
so r = 10
so for the angle in the first quadrant
sinØ = 8/10 = 4/5
cosØ = 6/10 = 3/5
tanØ = 8/6 = 4/3
for the point (-5,12) the triangle is in the 2nd quadrant,
x = -5, y = 12, and r = 13 , (r is considered always positive)
sinØ = 12/13
cosØ = -5/13
tanØ = -12/5
no matter what point you are given, plot the point and you will see which quadrant you are in.
Draw the perpendicular to the x-axis (not the y-axis) and follow the steps I outlined.
I don'tquite understand we never learned y and r stuff. We just started this..the chapter is called Angles of rotation. Andwhere did u get the 10 from... and um sorry for bothering you but can u help me do one because there are many..:(
Thanks I greatly appreciate you help!
join to the origin (0,0)
from (6,8) draw a perpendicular to the x-axis
to get a right-angled triangle , with the line from the origin to the point (6,8) as the hypotenuse.
If you are studying trig you must have learned the Pythagorean Theorem
let the length of the hypotenuse be r
then r^2 = 6^2 + 8^2
r^2 = 36+64 = 100
r = √100 = 10
if you label the rotated angle Ø,
sinØ = opposite/hypotenuse = 8/10 = 4/5
etc, see above
notice for the point (6,8), the 6 is the x value, and the 8 is the y value.
I hope that they will teach you very soon the trig definitions in terms of x, y , and r