How do I solve this: Tan(x) = -15/8, sin(x) > 0

Question

I tried putting in the Q2, but 15 was positive and 8 was negative. I tried putting it in Q4, but I could not get the right answer.

Answer 1

Since sin(x) > 0, x must be in either Quadrant I or II.

Since tan(x) < 0, x must be in Quadrant II or IV.

Thus, x must be in Quadrant II.

Because tan(x) = -15/8, and the tangent is equal to the y-coordinate divided by the x-coordinate, x must equal -8 OR some value such that its y-coordinate will reduce to -15/8 when divided by its x-coordinate.

Answer 2

tanT = -15/8 with sinT>0

T is in quadrant II

arcTan(-15/8) = about -61.93

T = arcTan(-15/8) + 180

T= about 118.07 degrees

= 118.07 x pi/180 radians= about 118pi/180

= about 59pi/90 radians

Answer 3

Draw a sketch:

I used θ_ref in the triangle (stands for the reference angle) but the real angle is not in the triangle but from the positive x-axis around the the hypotenuse of the triangle (shown as θ in blue).

Using your calculator (I'm going to use radians), type in tan^-1(-15/8) and you'll get -1.0808. This angle is in Q4 and the reason for your calculator giving you this answer is because the inverse tangent function give angles between -π/2 and π/2. You must translate your result to the appropriate Q2 position by adding π to it. x = π + -1.0808 = 2.0608. This is the angle in Q2.

Answer 4

desmos.com/calculator/gebgfh8ljl