Question

9. Given the point (6,-8) values of the six trig function.

10. Given that cot theta = -r3/2 in Quad II, find the state the six trig ratios. ​

Answer 1

Here's what I get.

Explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10

`\sin \theta = (-8)/(10) = -(4)/(5)\\\\\cos \theta =(6)/(10) = (3)/(5)\\\\\tan \theta = (-8)/(6) = -(4)/(3)\\\\\csc \theta = (10)/(-8) = -(5)/(4)\\\\\sec \theta = (10)/(6) = (5)/(3)\\\\\cot \theta = (6)/(-8) = -(3)/(4)`

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

`\sin \theta = (2)/(√(7)) = (2√(7))/(7)\\\\\cos \theta = (-√(3))/(√(7)) = -(√(21))/(7)\\\\\tan \theta = (2)/(-√(3)) = -(2√(3))/(3)\\\\\csc \theta = (√(7))/(2) \\\\\sec \theta = (√(7))/(-√(3)) = -(√(21))/(3)\\\\\cot \theta = -(√(3))/(2)`