The Unit Circle and "All Students Take Calculus"

Trigonometry is like a workout for the brain ;)
Some trig problems are posed like this:
sin A = 8/17 and cos A <0, find tan A.
How to start?

Where is sin positive and cos negative? Sin is +ve (positive) in Quadrants I and II. Cos is positive in Quadrant I so we must be in Quadrant II. The Quadrants are labelled with Roman numerals in a counterclockwise fashion starting with the most popular and +ve Quadrant #1.

In Quadrant II, sin is +ve but cos is -ve -- this results in a -ve tan since tan = sin/cos.

Tan = opp/adj

In Quadrant 2, we draw the right triangle of 8, 15, 17. Since sin A is 8/17 and the angle is in Quadrant II, we know that the y value (vertical part of the triangle) is positive. This makes the x value have to be -15. We also know this because in Quadrant II on a regular x, y coordinate plane, x is -ve while y is +ve.

From the angle, since tan = opp/adj and opp = 8 and adj = -15, then tan A = -8/15.

Note: We do not need to use a calculator for this!! (nor should we as the calculator will find the Quadrant I angle (less than 90) that pertains to sin A = 8/17 rather than the second angle value (between 90 and 180) where sin A also is equal to 8/17 but cos A <0.