Real-World Modeling with Inverse Sine

We are told the vertical component Tᵥ = T sin θ.

So, sin θ = Tᵥ / T = 400 / 800 = 1/2. To find θ, take the inverse sine: θ = arcsin(1/2). We know sin 30° = 1/2, so θ = 30°.

The formula is:

s = 5 sin θ

So, first Substitute s = 0

0 = 5 sin θ

Then, Divide both sides by 5

sin θ = 0

and finally, find the principal-value angle whose sine is 0

The principal range for arcsin is:

−90° ≤ θ ≤ 90°

In this interval, the only angle with sin θ = 0 is:

θ = 0°

The formula is: y = 2 sin(kx)

So, first Substitute y = −1

−1 = 2 sin(kx)

Then, Divide both sides by 2

sin(kx) = −1/2

And finally, find the principal-value angle whose sine is −1/2

So, The principal range for arcsin is:

−π/2 ≤ kx ≤ π/2

In this interval, the angle with sin(kx) = −1/2 is:

kx = −π/6

The formula is: vᵧ = v sin θ

So, first substitute the values:

6 = 12 sin θ

Then, divide both sides by 12:

sin θ = 6 / 12

sin θ = 1/2

And finally, find the angle whose sine is 1/2:

θ = 30°

So the skateboarder’s launch angle is 30°.

We know sin(π/6) = 1/2. Since 1/2 is positive, θ is in Quadrant I. Hence, θ = π/6.

The formula is: h = 1.2 + 0.8 sin(kt)

So, first substitute h = 1.2:

1.2 = 1.2 + 0.8 sin(kt)

Then, subtract 1.2 from both sides:

0 = 0.8 sin(kt)

Next, divide both sides by 0.8:

sin(kt) = 0

And finally, find the principal-value angle whose sine is 0:

kt = arcsin(0)

The formula is: y = 0.12 sin(ωt)

So, first substitute y = 0.06:

0.06 = 0.12 sin(ωt₀)

Then, divide both sides by 0.12:

sin(ωt₀) = 0.06 / 0.12

sin(ωt₀) = 0.5

And finally, solve for the angle:

ωt₀ = arcsin(0.5)

We are given: sin(18°) = h / 250

So, first multiply both sides by 250 to solve for h:

h = 250 · sin(18°)

This expression gives the height of the object above the horizontal line of sight.

The formula is: H = 30 sin θ

So, first, Substitute H = 24

24 = 30 sin θ

Then, Divide both sides by 30

sin θ = 24 / 30

sin θ = 0.8

And finally, find the principal-value angle whose sine is 24/30

To isolate θ, we apply the inverse sine function:

θ = arcsin(24/30)

The formula is: y = 250 + 40 sin(0.1x)

So, first Substitute y = 270

270 = 250 + 40 sin(0.1x)

Then, Subtract 250 from both sides

20 = 40 sin(0.1x)

Next, Divide both sides by 40

sin(0.1x) = 20 / 40

sin(0.1x) = 1/2

And finally, find the principal-value angle whose sine is 1/2

The principal range for arcsin is:

−π/2 ≤ 0.1x ≤ π/2

In this interval, the angle with sin(0.1x) = 1/2 is:

0.1x = arcsin(1/2)

Since 1/2 = 20/40, this matches:

0.1x = arcsin(20/40)

We know the relationship is: sin θ = opposite / hypotenuse

So, first write the ratio:

sin θ = 1.2 / 10

Then, simplify the fraction:

sin θ = 0.12

And finally, solve for the angle using the inverse sine:

θ = arcsin(0.12)

This gives an angle θ that is approximately 0.12 radians.

The formula is: y = 3.5 sin(ωt)

So, first substitute y = 3.5:

3.5 = 3.5 sin(ωt)

Then, divide both sides by 3.5:

sin(ωt) = 1

And finally, find the angle (in the principal range) whose sine is 1:

The principal range is: −π/2 ≤ ωt ≤ π/2.

In this interval, the angle with sine 1 is:

ωt = π/2

We know a basic reference fact: sin(π/6) = 1/2

To get a negative sine value, the angle must be below the x-axis, so we take the negative of π/6:

θ = −π/6

The principal range for arcsin is:

−π/2 ≤ θ ≤ π/2

Since −π/6 lies in this interval and sin(−π/6) = −1/2, the correct principal value is:

θ = −π/6

We use the definition of sine in a right triangle: sin θ = opposite / hypotenuse

Here, the opposite side is 7.5 m (height on the wall), and the hypotenuse is 12 m (ladder length).

So, first write: sin θ = 7.5 /12

Then, solve for θ using the inverse sine:

θ = arcsin(7.5 /12)

This expression gives the angle θ in radians.

The formula is: p(t) = 0.8 sin(ωt)

So, first substitute p(t) = −0.4:

−0.4 = 0.8 sin(ωt)

Then, divide both sides by 0.8:

sin(ωt) = −0.4 / 0.8

sin(ωt) = −1/2

And finally, find the principal-value angle whose sine is −1/2:

ωt = arcsin(−1/2)

We know sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2], so:

ωt = −π/6

So ωt = arcsin(−1/2) gives the correct principal value.

The formula is: h = 18 + 15 sin θ

So, first substitute h = 30:

30 = 18 + 15 sin θ

Then, subtract 18 from both sides:

12 = 15 sin θ

Next, divide both sides by 15:

sin θ = 12 / 15

sin θ = 4/5

And finally, solve for θ:

θ = arcsin(4/5)

We can form a right triangle with:

vertical (opposite) = 90 m

horizontal (adjacent) = 120 m

First, find the hypotenuse using Pythagoras:

hypotenuse = √(90² + 120²) = √(8100 + 14400) = √22500 = 150

Then, use the sine ratio:

sin θ = opposite / hypotenuse = 90 / 150

Simplify:

sin θ = 0.6

Now find the angle whose sine is 0.6:

θ = arcsin(0.6) ≈ 36.9°

Rounded to the nearest degree:

θ ≈ 37°

The slope 1:12 means: rise = 1, run = 12

We can model this with a right triangle where:

opposite = 1

hypotenuse is approximately 12 for small angles, or we use the approximation sin θ ≈ rise/run for gentle slopes.

So we take:

sin θ ≈ 1 / 12 ≈ 0.0833

Then find the angle whose sine is 0.0833:

θ = arcsin(1/12) ≈ 4.78°

Rounded to the nearest tenth:

θ ≈ 4.8°

The formula is: h = 2 + 2 sin θ

So, first substitute h = 3:

3 = 2 + 2 sin θ

Then, subtract 2 from both sides:

1 = 2 sin θ

Next, divide both sides by 2:

sin θ = 1 / 2

And finally, solve for θ:

θ = arcsin(1/2)

We interpret rise and run as sides of a right triangle: rise (opposite) = 24

hypotenuse is determined by the slope, but for the angle θ we use:

sin θ = opposite / hypotenuse

However, in many ramp problems, we model sin θ using rise over the length directly; here, the intended relationship is:

sin θ = 24 / 60 = 0.4

So, solve for θ:

θ = arcsin(24 / 60)