Two vectors of 10 units & 5 units make an angle of `120^(@)` with each other.Find the magnitude & angle of resultant with vector of 10 unit magnitude.

Answer

To solve the problem of finding the magnitude and angle of the resultant vector when two vectors of 10 units and 5 units make an angle of 120 degrees with each other, we can follow these steps:

### Step 1: Understand the Vectors
Let vector A be the vector with a magnitude of 10 units, and vector B be the vector with a magnitude of 5 units. The angle between them is given as 120 degrees.

### Step 2: Use the Law of Cosines to Find the Magnitude of the Resultant Vector
The magnitude of the resultant vector \( R \) can be calculated using the law of cosines:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
Where:
- \( A = 10 \) units
- \( B = 5 \) units
- \( \theta = 120^\circ \)

Substituting the values:
\[ R = \sqrt{10^2 + 5^2 + 2 \cdot 10 \cdot 5 \cdot \cos(120^\circ)} \]

### Step 3: Calculate \( \cos(120^\circ) \)
The cosine of 120 degrees is:
\[ \cos(120^\circ) = -\frac{1}{2} \]

### Step 4: Substitute \( \cos(120^\circ) \) into the Equation
Now substituting \( \cos(120^\circ) \) into the equation:
\[ R = \sqrt{10^2 + 5^2 + 2 \cdot 10 \cdot 5 \cdot \left(-\frac{1}{2}\right)} \]
\[ = \sqrt{100 + 25 - 50} \]
\[ = \sqrt{75} \]

### Step 5: Calculate the Magnitude of the Resultant
Thus, the magnitude of the resultant vector \( R \) is:
\[ R = \sqrt{75} = 5\sqrt{3} \text{ units} \]

### Step 6: Find the Angle of the Resultant with Vector A
To find the angle \( \phi \) between the resultant vector \( R \) and vector \( A \), we can use the formula:
\[ \cos(\phi) = \frac{A^2 + R^2 - B^2}{2AR} \]
Substituting the known values:
\[ \cos(\phi) = \frac{10^2 + (\sqrt{75})^2 - 5^2}{2 \cdot 10 \cdot \sqrt{75}} \]
\[ = \frac{100 + 75 - 25}{20\sqrt{75}} \]
\[ = \frac{150}{20\sqrt{75}} = \frac{15}{2\sqrt{75}} \]

### Step 7: Simplify and Calculate \( \phi \)
Now, we can simplify:
\[ \cos(\phi) = \frac{15}{2 \cdot 5\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \]
Thus, \( \phi = 30^\circ \).

### Final Result
The magnitude of the resultant vector is \( 5\sqrt{3} \) units, and the angle of the resultant with the vector of 10 unit magnitude is \( 30^\circ \).

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