Find the exact values of the remaining five trigonometric functions of o satisfying the given conditions. cos θ=3/5, tan θ < 0

First, tan(θ) = sin(θ) / cos(θ), so if cos(θ) = 3/5 > 0 and tan(θ) < 0, then it follows that sin(θ) < 0.

Recall the Pythagorean identity:

sin²(θ) + cos²(θ) = 1

Then

sin(θ) = -√(1 - cos²(θ)) = -4/5

and so

tan(θ) = (-4/5) / (3/5) = -4/3

The remaining trig ratios are just reciprocals of the ones found already:

sec(θ) = 1/cos(θ) = 5/3

csc(θ) = 1/sin(θ) = -5/4

cot(θ) = 1/tan(θ) = -3/4

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

step-by-step explanation:

assume that you have 3 different side lengths.   apply the pythagorean theorem here:   square each of the two shortest sides; call them a and b, so that their squares are a² and b².   also square the longest side, calling the result c².

now, if a² + b² = c² is true, the triangle is a right triangle; otherwise it is not a right triangle.