SOLUTION: How to solve a problem like this: Find the exact value of each of the remaining trigonometric functions of theta...My problem is cos= -4/5 thata, in Q III. Also, How do I solve

Algebra ->  Trigonometry-basics -> SOLUTION: How to solve a problem like this: Find the exact value of each of the remaining trigonometric functions of theta...My problem is cos= -4/5 thata, in Q III. Also, How do I solve       Log On


   

Question 410966: How to solve a problem like this: Find the exact value of each of the remaining trigonometric functions of theta...My problem is cos= -4/5 thata, in Q III.
Also, How do I solve sin =5/13, 90 degrees< theta <180 degrees.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
  1. Draw a right triangle. It doesn't really matter how it looks.
  2. Pick one of the acute angles to be theta. It doesn't matter which angle. The sides of this angle will be the hypotenuse and one of the legs of the triangle.
  3. Since cos is adjacent over hypotenuse, label the hypotenuse as 5 and the adjacent side (the leg that is a side of theta is the "adjacent" side) as -4.
  4. Use the Pythagorean Theorem to find the third side. You should get 3.
  5. Since theta terminates in the third quadrant (Q III), and since both x's and y's are negative in Q III, label the third side of the triangle as -3.
  6. Now that you have values for all three sides, form the ratios for the other five trig functions.

For the second problem, do the same thing except:
  • The hypotenuse will be 13.
  • The -5 will go on the opposite side (which is the leg that is not one of the sides of theta) since sin is opposite over hypotenuse.

Notes:
  • The hypotenuse is always positive in these problems. Only the adjacent and/or opposite sides of theta can be negative.
  • If you are given tan or cot then if the angle terminates in
    • Q I, make both legs positive.
    • Q II, make the adjacent leg negative and the opposite leg positive.
    • Q III, make both legs negative.
    • Q IV, make the adjacent side positive and the opposite side negative.

P.S. In respoonse to the question in your "Thank you" note...
In the Pythagorean equation, the hypotenuse is always by itself on one side of the equaiton and the legs are on the other side. So the equation to to find the 2nd leg when one leg is 4 and the hypotenuse is 5 would be:
x%5E2+%2B+4%5E2+=+5%5E2
which simplifies to:
x%5E2+%2B+16+=+25
Subtracting 16:
x%5E2+=+9
Square root:
x = 3
With the second problem you again know one leg and the hypotenuse. So the Pythagorean equation for that problem will be similar to this one.