Question 253677: I need help with a very difficult math assignment i have recieved my math teachers makes it way more complicated then it has to be.
1.) if sin x= (12/13), cos y=(3/5) and x and y are acute angles, the value of cos(x-y) is
a.(21/65) b.(63-65) c. -(14/65) d. -(33/65)
2.) if the tangent of an angle is negative and its secant is positive, in which quadrant does the angle terminate
1,2,3 or 4 i thought it was the second quadrant but i dont understand the term terminate i wasnt sure if that ment the opposite of what it was.
Found 2 solutions by drk, stanbon: Answer by drk(1908) (Show Source): Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 1.) if sin x= (12/13), cos y=(3/5) and x and y are acute angles, the value of cos(x-y) is
a.(21/65) b.(63/65) c. -(14/65) d. -(33/65)
---
cos(x-y) = cos(x)cos(y)+sin(x)sin(y)
---
Since sin(x) = 12/13, cos(x)=5/13
Since cos(y) = 3/5, sin(y) = 4/5
----
Therefore:
cos(x-y) = (5/13)(3/5)+(12/13)(4/5) = (63/65)
========================================================
2.) if the tangent of an angle is negative and its secant is positive, in which quadrant does the angle terminate
---
tangent is negative in the 2nd and 4th quadrants
secant is positive in the 1st and 4th quadrants.
Therefore the angle is in the 4th quadrant.
=========================================================
Cheers,
Stan H.
|
|
|