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 \(\vec{B}\) with magnitudes of 10 units each and angles of \(30^\circ\) and \(60^\circ\) with the x-axis respectively, we can follow these steps:

### Step 1: Determine the components of each vector
The components of a vector can be found using trigonometric functions. For vector \(\vec{A}\) and \(\vec{B}\):

- The x-component of \(\vec{A}\):
\[ A_x = A \cos(30^\circ) = 10 \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

- The y-component of \(\vec{A}\):
\[ A_y = A \sin(30^\circ) = 10 \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \]

- The x-component of \(\vec{B}\):
\[ B_x = B \cos(60^\circ) = 10 \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \]

- The y-component of \(\vec{B}\):
\[ B_y = B \sin(60^\circ) = 10 \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

### Step 2: Calculate the resultant components
Now, we can find the resultant vector \(\vec{R}\) by summing the components of \(\vec{A}\) and \(\vec{B}\):

- Resultant x-component:
\[ R_x = A_x + B_x = 5\sqrt{3} + 5 \]

- Resultant y-component:
\[ R_y = A_y + B_y = 5 + 5\sqrt{3} \]

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

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

Now, adding these:
\[ R^2 = (100 + 50\sqrt{3}) + (100 + 50\sqrt{3}) = 200 + 100\sqrt{3} \]

Finally, taking the square root:
\[ R = \sqrt{200 + 100\sqrt{3}} \]

### Step 4: Approximate the value
Using \(\sqrt{3} \approx 1.732\):
\[ R \approx \sqrt{200 + 100 \cdot 1.732} = \sqrt{200 + 173.2} = \sqrt{373.2} \]

Thus, the resultant vector \(\vec{R}\) is approximately:
\[ R \approx 19.3 \text{ units} \]

### Final Answer
The magnitude of the resultant vector \(\vec{R}\) is approximately \(19.3\) units.