Let `vecA and vecB` be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles `30^@ and 60^@` respectively, find the resultant.

Explanation

To find the resultant of the two vectors \(\vec{A}\) and \(\vec{B}\), we will follow these steps:

### Step 1: Identify the Magnitudes and Angles
Given:
- Magnitude of \(\vec{A} = 10\) units (inclined at \(30^\circ\) to the X-axis)
- Magnitude of \(\vec{B} = 10\) units (inclined at \(60^\circ\) to the X-axis)

### Step 2: Resolve the Vectors into Components
We need to resolve both vectors into their X and Y components.

For vector \(\vec{A}\):
- \(A_x = A \cos(30^\circ) = 10 \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\)
- \(A_y = A \sin(30^\circ) = 10 \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5\)

For vector \(\vec{B}\):
- \(B_x = B \cos(60^\circ) = 10 \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5\)
- \(B_y = B \sin(60^\circ) = 10 \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\)

### Step 3: Calculate the Resultant Components
Now, we will add the components of both vectors to find the resultant vector \(\vec{R}\).

- \(R_x = A_x + B_x = 5\sqrt{3} + 5\)
- \(R_y = A_y + B_y = 5 + 5\sqrt{3}\)

### Step 4: Calculate the Magnitude of the Resultant Vector
The magnitude of the resultant vector \(\vec{R}\) can be calculated using the Pythagorean theorem:

\[ R = \sqrt{R_x^2 + R_y^2} \]

Substituting the values:

\[ R = \sqrt{(5\sqrt{3} + 5)^2 + (5 + 5\sqrt{3})^2} \]

### Step 5: Simplify the Expression
Calculating \(R_x^2\) and \(R_y^2\):

1. \(R_x^2 = (5\sqrt{3} + 5)^2 = 25 \cdot 3 + 50\sqrt{3} + 25 = 75 + 50\sqrt{3} + 25 = 100 + 50\sqrt{3}\)
2. \(R_y^2 = (5 + 5\sqrt{3})^2 = 25 + 50\sqrt{3} + 75 = 100 + 50\sqrt{3}\)

Now, adding these:

\[ R^2 = (100 + 50\sqrt{3}) + (100 + 50\sqrt{3}) = 200 + 100\sqrt{3} \]

Thus,

\[ R = \sqrt{200 + 100\sqrt{3}} \]

### Step 6: Final Result
The magnitude of the resultant vector \(\vec{R}\) is:

\[ R = \sqrt{200 + 100\sqrt{3}} \]