Using Clues to Find Trig Ratios

The tangent is positive in QI and in QIII. The sine ratio is negative in QIII and QIV. The overlap is QIII, so α must terminate in the third quadrant.

In the third quadrant, each of x and y is negative, so the numerator and denominator of the tangent ratio y/x are both negative.

I'll draw what I've got so far:

The Pythagorean Theorem gives me:

The hypotenuse r is always positive, so r = +4 and now my picture becomes:

This picture contains all the information that I need in order to find the other ratios. So those other ratios have the following values:

This is an unusual sort of exercise, but I can use what I've learned in algebra to pick it apart.

First, I'll solve for "y=" to get the equation of the line:

From what I remember of graphing, I know that this is an decreasing line through the origin, so it passes through QII and QIV. Since they gave me the restriction that x is negative, then the angle alpha must end in QII.

To get a point on this line, I'll plug −3 for x, so I get 5 for y (and whole-number coordinates for easy graphing and computation).

So I've got this much so far:

Dropping the perpendicular to create the right triangle, I can label the side lengths from the point, and also label the angle, to get:

The height is y = 5 and the base is x = −3, so the hypotenuse is given by:

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Since r is always positive, then the hypotenuse takes the positive value, , and my triangle is:

Then the six trig ratios are: