What is the tangent for an angle that has a cosine of −17√ and is in Quadrant III? Use the Pythagorean identity sin2(θ)+cos2(θ)=1 , the trigonometric identity tanθ=sinθcosθ , and the quadrant to solve. (1 point)
Responses
6
6
−6
negative 6
−6–√
negative Start Root 6 End Root
6–√
The tangent of an angle in Quadrant III can be found using the trigonometric identity tanθ=sinθ/cosθ.
First, we need to find the sine of the angle. Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we can solve for sinθ.
sin^2(θ) + cos^2(θ) = 1
sin^2(θ) = 1 - cos^2(θ)
sinθ = √(1 - cos^2(θ))
sinθ = √(1 - (-17√)^2)
sinθ = √(1 - 289)
sinθ = √(1 - 289)
sinθ = √(-288)
Since the angle is in Quadrant III, sinθ is negative. Therefore, sinθ = -√288.
Now we can find the tangent of the angle using tanθ = sinθ/cosθ.
tanθ = (-√288)/(-17√)
tanθ = √288/17√
tanθ = (12√2)/(√17)(√)
tanθ = (12√2)/(√17)
Since the angle is in Quadrant III, the tangent is negative. Therefore, the tangent of the angle is - (12√2)/(√17).