Suppose cos A = − 3/5 , and π < A < 3 π 2 . Find sin A and tan A .

Answer 1

Hi Neven P.,

If cos(A) = -3/5 and the parameters are π < A < 3π/2, then that puts the triangle in the 3rd quadrant of the Cartesian coordinates. In the 3rd quadrant x-values and y-values are negative, so both our side opposite and side adjacent to angle A are going to be negative. This is exactly why cos(A) is negative (-3/5) because the x-value is negative (cos (A) = x/H).

Since cos(A) = -3/5 the triangles x value is -3 (side adjacent), and the hypotenuse is 5. This forms a 3,4,5 right triangle in quadrant 3, but you can also use the Pythagorean theorem to solve for the y value (side opposite), y = ±√(5² - 3²) = -4, and we choose the negative y-value because we are in quadrant 3.

We have all the sides of the triangle:

x, or side adjacent (SA) = -3;

y, or side opposite (SO) = -4

Hypotenuse (H) = 5

Then for:

sin(A) = y/H = SO/H = -4/5

tan(A) = y/x = SO/SA = -4/-3 = 4/3

Notice that sin(A) and cos(A) are negative in quadrant 3, and tan(A) is positive.

I hope this helps, Joe.

Answer 2

Is 3π2 supposed to be 3π^2?

acos-.6=.1268........