Physics worksheet on vector addition, showing vector diagrams and component calculations.

Explanation:
We are given several vectors with magnitude and direction (angle), and we need to find their x- and y-components. Then, we add vectors by adding their components separately.

Recall:
For a vector with magnitude V and angle θ measured from the positive x-axis (counterclockwise is positive), the components are:

- $ x = V \cos\theta $
- $ y = V \sin\theta $

⚠️ Important: Angles must be measured from the +x-axis, *not* from the axes in the diagram unless specified. In many diagrams like these, angles shown are relative to the nearest axis — so we must convert them to standard position (from +x-axis, CCW positive).

Let’s go step by step.

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First row: Find x and y components

1. Vector: 12 m at 30° above +x-axis
→ θ = 30°
$ x = 12 \cos 30^\circ = 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.39 $
$ y = 12 \sin 30^\circ = 12 \cdot \frac{1}{2} = 6 $

2. Vector: 25 m at 60° above +x-axis
→ θ = 60°
$ x = 25 \cos 60^\circ = 25 \cdot \frac{1}{2} = 12.5 $
$ y = 25 \sin 60^\circ = 25 \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{2} \approx 21.65 $

3. Vector: 15 m at 150° (given as 150° — this is already from +x-axis, CCW)
→ θ = 150°
$ \cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} $
$ \sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2} $
So:
$ x = 15 \cdot (-\frac{\sqrt{3}}{2}) = -\frac{15\sqrt{3}}{2} \approx -12.99 $
$ y = 15 \cdot \frac{1}{2} = 7.5 $

So first row answers:
- First vector: x ≈ 10.39 m, y = 6 m
- Second: x = 12.5 m, y ≈ 21.65 m
- Third: x ≈ −12.99 m, y = 7.5 m

But since the worksheet likely expects exact forms or decimals rounded reasonably, and often uses √3 ≈ 1.732, let’s keep 2 decimal places unless otherwise noted.

Now second part: Add vectors

First addition:
Vector A: 12 m at 20° above +x-axis
→ $ x_1 = 12 \cos 20^\circ $, $ y_1 = 12 \sin 20^\circ $
Vector B: 12 m at 70° above +y-axis? Wait — look carefully: the angle is labeled 70° between vector and vertical (y-axis), on the left side of y-axis → that means it's in the second quadrant, 70° from +y toward −x.

So angle from +x-axis = 90° + 70° = 160°
(From +x, go 90° to +y, then another 70° toward −x → total 160°)

Thus:
- $ x_2 = 12 \cos 160^\circ $
- $ y_2 = 12 \sin 160^\circ $

Compute:

$ \cos 20^\circ \approx 0.9397 $ → $ x_1 = 12 × 0.9397 ≈ 11.28 $
$ \sin 20^\circ \approx 0.3420 $ → $ y_1 = 12 × 0.3420 ≈ 4.10 $

$ \cos 160^\circ = \cos(180−20) = -\cos 20^\circ ≈ -0.9397 $ → $ x_2 ≈ 12 × (-0.9397) = -11.28 $
$ \sin 160^\circ = \sin(180−20) = \sin 20^\circ ≈ 0.3420 $ → $ y_2 ≈ 4.10 $

Now sum:
- $ x_{\text{net}} = 11.28 + (-11.28) = 0 $
- $ y_{\text{net}} = 4.10 + 4.10 = 8.20 $

So result: x = 0, y ≈ 8.20 m
(Exact: $ x = 12(\cos20^\circ - \cos20^\circ) = 0 $, $ y = 12(\sin20^\circ + \sin20^\circ) = 24\sin20^\circ $)

Now second addition:

Vector C: 50 m at 135° (shown as 135° from +x-axis — standard position, since it's in QII, 135° is correct)
→ $ x_1 = 50 \cos 135^\circ $, $ y_1 = 50 \sin 135^\circ $

Vector D: 30 m straight down → that’s 270° from +x-axis, or −90°
→ $ x_2 = 30 \cos 270^\circ = 30 × 0 = 0 $
→ $ y_2 = 30 \sin 270^\circ = 30 × (-1) = -30 $

Now compute C:
- $ \cos 135^\circ = \cos(180−45) = -\cos 45^\circ = -\frac{\sqrt{2}}{2} ≈ -0.7071 $
- $ \sin 135^\circ = \sin(180−45) = \sin 45^\circ = \frac{\sqrt{2}}{2} ≈ 0.7071 $

So:
- $ x_1 = 50 × (-0.7071) ≈ -35.36 $
- $ y_1 = 50 × 0.7071 ≈ 35.36 $

Add:
- $ x_{\text{net}} = -35.36 + 0 = -35.36 $
- $ y_{\text{net}} = 35.36 + (-30) = 5.36 $

So final net vector: x ≈ −35.36 m, y ≈ 5.36 m

Now, the worksheet asks to fill in blanks for each component before and after addition.

Let’s write all required values clearly, using two decimal places (standard for physics worksheets unless exact radicals requested — but here likely decimals).

First row (components only):
1. 12 m @ 30°:
x = 12 cos30° = 10.39 m
y = 12 sin30° = 6.00 m

2. 25 m @ 60°:
x = 25 cos60° = 12.50 m
y = 25 sin60° = 21.65 m

3. 15 m @ 150°:
x = 15 cos150° = −12.99 m
y = 15 sin150° = 7.50 m

Second part — first addition:
Vector 1: 12 m @ 20°
x₁ = 11.28 m
y₁ = 4.10 m

Vector 2: 12 m @ 160° (since 70° from +y toward −x)
x₂ = −11.28 m
y₂ = 4.10 m

Sum:
x_net = 0.00 m
y_net = 8.20 m

Third part (second addition):
Vector 1: 50 m @ 135°
x₁ = −35.36 m
y₁ = 35.36 m

Vector 2: 30 m ↓ (270°)
x₂ = 0.00 m
y₂ = −30.00 m

Sum:
x_net = −35.36 m
y_net = 5.36 m

Let me double-check all calculations:

- cos30 = √3/2 ≈ 0.8660 → 12×0.8660 = 10.392 → 10.39 ✓
- sin30 = 0.5 → 6.00 ✓
- cos60 = 0.5 → 12.5 ✓
- sin60 = √3/2 ≈ 0.8660 → 25×0.8660 = 21.65 ✓
- cos150 = −√3/2 ≈ −0.8660 → 15×−0.8660 = −12.99 ✓
- sin150 = 0.5 → 7.5 ✓

- cos20 ≈ 0.9396926 → 12× = 11.2763 ≈ 11.28 ✓
- sin20 ≈ 0.3420201 → 12× = 4.1042 ≈ 4.10 ✓
- cos160 = cos(180−20) = −cos20 ≈ −0.9397 → −11.28 ✓
- sin160 = sin20 ≈ 0.3420 → 4.10 ✓
→ Sum x = 0, y = 8.20 ✓

- cos135 = −√2/2 ≈ −0.7071068 → 50× = −35.355 ≈ −35.36 ✓
- sin135 = √2/2 ≈ 0.7071 → 35.355 ≈ 35.36 ✓
- y: 35.355 − 30 = 5.355 ≈ 5.36 ✓

All good.

Final Answer:
First row:
x = 10.39 m, y = 6.00 m
x = 12.50 m, y = 21.65 m
x = −12.99 m, y = 7.50 m

Second addition:
x₁ = 11.28 m, y₁ = 4.10 m
x₂ = −11.28 m, y₂ = 4.10 m
x_net = 0.00 m, y_net = 8.20 m

Third addition:
x₁ = −35.36 m, y₁ = 35.36 m
x₂ = 0.00 m, y₂ = −30.00 m
x_net = −35.36 m, y_net = 5.36 m