Question

Angle θ is in standard position and terminates in quadrant II. If tanθ = -15/8, then cscθ = _____.

Answer 1

Answer:

csc Θ = $\frac{17}{15}$

Step-by-step explanation:

Given

tan Θ = - $\frac{15}{8}$ = $\frac{opposite}{adjacent}$ ← of a right triangle , then

hypotenuse h² = 15² + 8² = 225 + 64 = 289 ( square root both sides )

h = $\sqrt{289}$ = 17

Using the identity

csc x = $\frac{1}{sinx}$

sin Θ = $\frac{opposite}{hypotenuse}$ = $\frac{15}{17}$

Since Θ is in second quadrant then sinΘ and csc Θ > 0, thus

csc Θ = $\frac{1}{\frac{15}{17} }$ = $\frac{17}{15}$