SOH-CAH-TOA — the foundation of vector components and 2D physics
| Name | Ratio | Mnemonic |
|---|---|---|
| Sine | sin θ = opposite / hypotenuse | SOH |
| Cosine | cos θ = adjacent / hypotenuse | CAH |
| Tangent | tan θ = opposite / adjacent | TOA |
| Triangle | Angles | Side Ratios | sin/cos/tan Values |
|---|---|---|---|
| 45°-45°-90° | 45°, 45°, 90° | 1 : 1 : √2 | sin45°=cos45°=√2/2, tan45°=1 |
| 30°-60°-90° | 30°, 60°, 90° | 1 : √3 : 2 | sin30°=½, cos30°=√3/2, tan30°=1/√3 sin60°=√3/2, cos60°=½, tan60°=√3 |
When you know sides and need the angle:
A 10 kg block sits on a 30° incline. Find the components of weight parallel and perpendicular to the plane.
Weight W = mg = 100 N. Angle with perpendicular = 30°.
W_parallel = W sin30° = 100 × 0.5 = 50 N (down the incline)
W_perp = W cos30° = 100 × (√3/2) = 86.6 N (into the plane)
A projectile has initial velocity components v_x = 20 m/s, v_y = 15 m/s. Find launch angle and speed.
tan θ = v_y / v_x = 15/20 = 0.75
θ = tan⁻¹(0.75) = 36.9°
v₀ = √(v_x² + v_y²) = √(400 + 225) = √625 = 25 m/s
5 kg block on 25° incline, μ = 0.3. Find acceleration.
W = 50 N
W_∥ = 50 sin25° = 21.1 N
W_⊥ = 50 cos25° = 45.3 N = N (normal force)
f = μN = 0.3 × 45.3 = 13.6 N
F_net = W_∥ - f = 21.1 - 13.6 = 7.5 N
a = F_net/m = 7.5/5 = 1.5 m/s²
1) opp=5, adj=5√3≈8.66 2) hyp=10, angles≈36.9°, 53.1° 3) F_x=40 N, F_y=30 N
4) v_x=v_y=21.2 m/s 5) W_∥=34.2 N, W_⊥=94.0 N 6) R=50 N, θ=53.1° 7) height=4.33m, base=2.5m
8) cos θ=0.8, tan θ=0.75 (3-4-5 triangle!)