Question

1. Given the ratio below, find the other two trig ratios for it. Show your work. Make sure to give a rationalized answer.

1) tan θ = (3-√3)/9

2. Find the other two trig ratios for it. Show your work.
2) sin θ = 8/10

Answer 1

Explanation:

1) We can start by finding the adjacent and opposite sides of the right triangle using the Pythagorean theorem. Let x be the length of the adjacent side, then:

tan θ = (3-√3)/9

tan θ = opposite/adjacent

(3-√3)/9 = opposite/x

opposite = (3-√3)x/9

Using Pythagorean theorem:

(adjacent)^2 + (opposite)^2 = (hypotenuse)^2

x^2 + [(3-√3)x/9]^2 = (hypotenuse)^2

x^2 + (9-6√3+3)x^2/81 = (hypotenuse)^2

(82-54√3)x^2/81 = (hypotenuse)^2

We can simplify this expression to find the hypotenuse:

(hypotenuse)^2 = [(82-54√3)/81]x^2

hypotenuse = sqrt[(82-54√3)/81]x

Now we can find the other trig ratios:

cos θ = adjacent/hypotenuse = x/sqrt[(82-54√3)/81]x

cos θ = 1/sqrt[(82-54√3)/81]

cos θ = sqrt[(82+54√3)/81] / [(82-54√3)/81]

cos θ = sqrt(82+54√3) / (82-54√3)

sin θ = opposite/hypotenuse = (3-√3)x/9 / sqrt[(82-54√3)/81]x

sin θ = (3-√3) / sqrt(82-54√3)

2) To find the other trig ratios, we need to first find the adjacent and hypotenuse sides of the right triangle. Let x be the length of the adjacent side, then:

sin θ = opposite/hypotenuse = 8/10

opposite = (8/10)hypotenuse = 0.8hypotenuse

Using Pythagorean theorem:

(adjacent)^2 + (opposite)^2 = (hypotenuse)^2

x^2 + (0.8hypotenuse)^2 = hypotenuse^2

x^2 + 0.64h^2 = h^2

x^2 = 0.36h^2

x = 0.6h

Now we can find the other trig ratios:

cos θ = adjacent/hypotenuse = 0.6h/h = 0.6

tan θ = opposite/adjacent = (8/10)/0.6 = 4/3

cot θ = 1/tan θ = 3/4

Answer 2

1) We can start by using the Pythagorean identity to find the value of cos θ:

cos^2 θ + sin^2 θ = 1

sin^2 θ = 1 - cos^2 θ

tan θ = sin θ / cos θ

(3-√3)/9 = sin θ / cos θ

We can now substitute sin^2 θ = 1 - cos^2 θ into the equation:

(3-√3)/9 = (sqrt(1-cos^2 θ))/cos θ

(3-√3)/9 = (cos^-1(θ))/cos θ

(3-√3)/9 = (1/cos θ) - cos θ

(3-√3)/9 + cos θ = 1/cos θ

cos θ [(3-√3)/9 + cos θ] = 1

cos θ = 1 / [(3-√3)/9 + cos θ]

cos θ = 1 / [(3-√3)/9 + (3+√3)/9]

cos θ = 1 / (2√3/9)

cos θ = (9/2√3)

Now, we can use the other trigonometric ratios:

sin θ = tan θ cos θ

sin θ = [(3-√3)/9] [(9/2√3)]

sin θ = (3-√3)/6

csc θ = 1/sin θ

csc θ = 1/[(3-√3)/6]

csc θ = (6/(3-√3))

2) We can use the Pythagorean identity to find the value of cos θ:

cos^2 θ + sin^2 θ = 1

cos^2 θ = 1 - sin^2 θ

We know that sin θ = 8/10 = 4/5, so:

cos^2 θ = 1 - (4/5)^2

cos^2 θ = 1 - 16/25

cos^2 θ = 9/25

cos θ = ±3/5

Since sin θ is positive and θ is in the first or second quadrant, we can take cos θ = 3/5:

tan θ = sin θ / cos θ

tan θ = (4/5) / (3/5)

tan θ = 4/3

csc θ = 1/sin θ

csc θ = 1/(8/10)

csc θ = 5/4

sec θ = 1/cos θ

sec θ = 1/(3/5)

sec θ = 5/3

Hoped this helped!