Real-World Modeling With Inverse Sine

1) A cable exerts tension T = 800 N at angle θ, producing a vertical component Tᵥ = 400 N. Find θ in degrees.

22.5°

30°

45°

60°

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°.

Explanation

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°.

2) The vertical displacement of a roller-coaster car is s = 5 sin θ. If s = 0 in the principal range, what is θ?

θ = π/2

θ = 0

θ = arcsin(0)

Both B and C

Set s = 0 → 0 = 5 sin θ ⇒ sin θ = 0. Now θ = arcsin(0). The sine of 0 is 0, and 0 is within the range [−π/2, π/2]. Thus, θ = 0 or θ = arcsin(0) are equivalent expressions.

Explanation

Set s = 0 → 0 = 5 sin θ ⇒ sin θ = 0. Now θ = arcsin(0). The sine of 0 is 0, and 0 is within the range [−π/2, π/2]. Thus, θ = 0 or θ = arcsin(0) are equivalent expressions.

3) A boat moves along a wave modeled by y = 2 sin(kx). If y = −1 and kx ∈ [−π/2, π/2], find kx.

Kx = arcsin(−1/2)

Kx = −arcsin(2)

Kx = arcsin(1/2)

Kx = sin(−1/2)

Substitute y = −1 into the model: 2 sin(kx) = −1. Divide both sides by 2 → sin(kx) = −1/2. To find the angle, use kx = arcsin(−1/2). The negative sine indicates the angle is below the x-axis, within [−π/2, π/2].

Explanation

Substitute y = −1 into the model: 2 sin(kx) = −1. Divide both sides by 2 → sin(kx) = −1/2. To find the angle, use kx = arcsin(−1/2). The negative sine indicates the angle is below the x-axis, within [−π/2, π/2].

4) A stage light shines on a wall 30 m away. The height on the wall is H = 30 sin θ. If H = 24 m, find θ.

θ = sin(24/30)

θ = arcsin(30/24)

θ = arcsin(24)

θ = arcsin(24/30)

Substitute H = 24 → 24 = 30 sin θ. Divide by 30 → sin θ = 24/30 = 0.8. The angle is θ = arcsin(0.8) = arcsin(24/30). This gives θ ≈ 53.1°.

Explanation

Substitute H = 24 → 24 = 30 sin θ. Divide by 30 → sin θ = 24/30 = 0.8. The angle is θ = arcsin(0.8) = arcsin(24/30). This gives θ ≈ 53.1°.

5) A hillside’s elevation is y = 250 + 40 sin(0.1x). If y = 270, find 0.1x in [−π/2, π/2].

0.1x = arcsin(40/20)

0.1x = arcsin(270/40)

0.1x = sin(1/2)

0.1x = arcsin(20/40)

Substitute y = 270 → 270 = 250 + 40 sin(0.1x). Subtract 250 → 20 = 40 sin(0.1x). Divide by 40 → sin(0.1x) = 1/2. Hence, 0.1x = arcsin(1/2) = arcsin(20/40).

Explanation

Substitute y = 270 → 270 = 250 + 40 sin(0.1x). Subtract 250 → 20 = 40 sin(0.1x). Divide by 40 → sin(0.1x) = 1/2. Hence, 0.1x = arcsin(1/2) = arcsin(20/40).

6) A radio antenna 10 m long deflects 1.2 m horizontally. If sin θ = 1.2/10, find θ in radians.

θ = arcsin(0.12)

θ = 10·arcsin(0.12)

θ = arcsin(10/1.2)

θ = sin(0.12)

We know sin θ = opposite/hypotenuse = 1.2/10 = 0.12. Take the inverse sine to find the angle: θ = arcsin(0.12). This gives θ ≈ 0.12 radians.

Explanation

We know sin θ = opposite/hypotenuse = 1.2/10 = 0.12. Take the inverse sine to find the angle: θ = arcsin(0.12). This gives θ ≈ 0.12 radians.

7) A skateboarder launches off a ramp at angle θ. If vᵧ = v sin θ with v = 12 m/s and vᵧ = 6 m/s, find θ in degrees.

15°

30°

45°

60°

Substitute values: 6 = 12 sin θ ⇒ sin θ = 1/2. The angle with sine 1/2 is 30°. Hence, θ = 30°.

Explanation

Substitute values: 6 = 12 sin θ ⇒ sin θ = 1/2. The angle with sine 1/2 is 30°. Hence, θ = 30°.

8) A signal is y = 3.5 sin(ωt). If y = 3.5 and ωt ∈ [−π/2, π/2], find ωt.

ωt = π/2

ωt = arcsin(3.5)

ωt = arcsin(1)

ωt = 0

Substitute y = 3.5 → 3.5 = 3.5 sin(ωt). So sin(ωt) = 1. The angle whose sine is 1 is π/2. Therefore, ωt = π/2.

Explanation

Substitute y = 3.5 → 3.5 = 3.5 sin(ωt). So sin(ωt) = 1. The angle whose sine is 1 is π/2. Therefore, ωt = π/2.

9) For sin θ = −1/2 with θ in the principal range, find θ.

−π/6

−π/2

π/6

π/3

We know sin(π/6) = 1/2. For a negative sine value, the angle is below the x-axis. Thus, θ = −π/6 in the range [−π/2, π/2].

Explanation

We know sin(π/6) = 1/2. For a negative sine value, the angle is below the x-axis. Thus, θ = −π/6 in the range [−π/2, π/2].

10) A surfer’s ramp angle satisfies sin θ = 1/2. What is θ?

−π/6

0

π/6

π/2

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

Explanation

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

11) A buoy's height is h = 1.2 + 0.8 sin(kt). If h = 1.2, find kt.

Kt = arcsin(1.2/0.8)

Kt = arcsin(0)

Kt = 0

Both B and C

Substitute h = 1.2 → 1.2 = 1.2 + 0.8 sin(kt). Simplify → sin(kt) = 0. So kt = arcsin(0) = 0. Both expressions mean the same thing.

Explanation

Substitute h = 1.2 → 1.2 = 1.2 + 0.8 sin(kt). Simplify → sin(kt) = 0. So kt = arcsin(0) = 0. Both expressions mean the same thing.

12) A ladder leans against a wall making angle θ with the ground. If top reaches 7.5 m and ladder is 12 m, find θ in radians.

θ = arcsin(7.5/12)

θ = arcsin(12/7.5)

θ = sin(7.5/12)

θ = arcsin(7.5)

By definition, sin θ = opposite / hypotenuse = 7.5 / 12. Take the inverse sine: θ = arcsin(7.5/12). That gives the angle in radians.

Explanation

By definition, sin θ = opposite / hypotenuse = 7.5 / 12. Take the inverse sine: θ = arcsin(7.5/12). That gives the angle in radians.

13) The equation p(t) = 0.8 sin(ωt). If p(t) = −0.4, find ωt.

ωt = −arcsin(1/2)

ωt = arcsin(−1/2)

ωt = arcsin(1/2)

ωt = −sin(1/2)

Substitute p(t) = −0.4 → −0.4 = 0.8 sin(ωt). Divide by 0.8 → sin(ωt) = −1/2. Thus, ωt = arcsin(−1/2) = −π/6.

Explanation

Substitute p(t) = −0.4 → −0.4 = 0.8 sin(ωt). Divide by 0.8 → sin(ωt) = −1/2. Thus, ωt = arcsin(−1/2) = −π/6.

14) A Ferris wheel has h = 18 + 15 sin θ. If h = 30, find θ.

θ = arcsin(4/5)

θ = arcsin(2)

θ = arcsin(3/5)

θ = sin(2/3)

Substitute h = 30 → 30 = 18 + 15 sin θ. Simplify → 12 = 15 sin θ ⇒ sin θ = 4/5. Thus θ = arcsin(4/5) ≈ 0.93 radians.

Explanation

Substitute h = 30 → 30 = 18 + 15 sin θ. Simplify → 12 = 15 sin θ ⇒ sin θ = 4/5. Thus θ = arcsin(4/5) ≈ 0.93 radians.

15) A drone ascends at angle θ above horizontal. It gains 90 m altitude over 120 m horizontal. Find θ to nearest degree.

37°

49°

53°

60°

Form a right triangle with opposite = 90 and adjacent = 120. sin θ = 90 / √(90² + 120²) = 0.6. θ = arcsin(0.6) ≈ 36.9°, rounded to 37°.

Explanation

Form a right triangle with opposite = 90 and adjacent = 120. sin θ = 90 / √(90² + 120²) = 0.6. θ = arcsin(0.6) ≈ 36.9°, rounded to 37°.

16) An incline has slope 1:12 (rise:run). Find maximum angle θ in degrees.

2.4°

4.8°

3.8°

5.7°

sin θ = 1 / 12 ≈ 0.0833. θ = arcsin(0.0833) ≈ 4.78°. Rounded to the nearest tenth, θ ≈ 4.8°.

Explanation

sin θ = 1 / 12 ≈ 0.0833. θ = arcsin(0.0833) ≈ 4.78°. Rounded to the nearest tenth, θ ≈ 4.8°.

17) A wheel of radius 2 m rotates so h = 2 + 2 sin θ. If h = 3, find θ.

θ = arcsin(1/2)

θ = arcsin(3/2)

θ = arcsin(2/3)

θ = sin(1/2)

3 = 2 + 2 sin θ → sin θ = 1/2. The angle whose sine is 1/2 is π/6. Thus θ = arcsin(1/2) = π/6 ≈ 0.52 radians.

Explanation

3 = 2 + 2 sin θ → sin θ = 1/2. The angle whose sine is 1/2 is π/6. Thus θ = arcsin(1/2) = π/6 ≈ 0.52 radians.

18) A pendulum’s displacement is y = 0.12 sin(ωt). At time t₀, y = 0.06, find ωt₀.

ωt₀ = arcsin(0.06)

ωt₀ = arcsin(0.5)

ωt₀ = arcsin(2)

ωt₀ = sin(0.5)

Substitute y = 0.06 → 0.06 = 0.12 sin(ωt₀). Divide by 0.12 → sin(ωt₀) = 0.5. Hence, ωt₀ = arcsin(0.5) = π/6.

Explanation

Substitute y = 0.06 → 0.06 = 0.12 sin(ωt₀). Divide by 0.12 → sin(ωt₀) = 0.5. Hence, ωt₀ = arcsin(0.5) = π/6.

19) A ship’s sonar measures angle of elevation 18°. sin(18°) = h/250. Find h.

H = arcsin(250 · 18°)

H = 250 · arcsin(18°)

H = 250 · sin(18°)

H = sin(250/18°)

Multiply both sides by 250 → h = 250 sin(18°). This gives the height of the cliff directly.

Explanation

Multiply both sides by 250 → h = 250 sin(18°). This gives the height of the cliff directly.

20) A ramp rises 24 inches over 60 inches horizontally. Which expression gives θ in radians?

θ = arcsin(24/60)

θ = arcsin(60/24)

θ = arcsin(24/36)

θ = arcsin(36/60)

The sine of the ramp’s angle equals rise/run = 24/60 = 0.4. θ = arcsin(0.4) = arcsin(24/60). This gives the correct angle in radians.

Explanation

The sine of the ramp’s angle equals rise/run = 24/60 = 0.4. θ = arcsin(0.4) = arcsin(24/60). This gives the correct angle in radians.

Real-World Modeling with Inverse Sine