Two vectors `vec P and vec Q` are inclinded to each other at `60^(@)`. The magnitude of `vec P and vec Q` are 10 and 15 units respectively. What is the angle `(alpha)` made by their resultant `vec R" with "vec P` ?

Options

A. `alpha = tan^(-1)(0.5)`
B. `alpha = tan^(-1)(0.6134)`
C. `alpha = tan^(-1)(0.7422)`
D. `alpha = tan^(-1) (0.92)`

Explanation

To solve the problem, we need to find the angle \( \alpha \) made by the resultant vector \( \vec{R} \) with vector \( \vec{P} \). We will use the law of vector addition and some trigonometric identities.

### Step 1: Understand the given information
We have two vectors:
- \( \vec{P} \) with a magnitude of 10 units
- \( \vec{Q} \) with a magnitude of 15 units
- The angle \( \theta \) between them is \( 60^\circ \).

### Step 2: Use the formula for the tangent of the angle \( \alpha \)
The angle \( \alpha \) made by the resultant vector \( \vec{R} \) with vector \( \vec{P} \) can be found using the formula:
\[ \tan \alpha = \frac{|\vec{Q}| \sin \theta}{|\vec{P}| + |\vec{Q}| \cos \theta} \]
Where:
- \( |\vec{P}| = 10 \)
- \( |\vec{Q}| = 15 \)
- \( \theta = 60^\circ \)

### Step 3: Calculate \( \sin \theta \) and \( \cos \theta \)
Using trigonometric values:
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 60^\circ = \frac{1}{2} \)

### Step 4: Substitute the values into the formula
Substituting the values into the formula gives:
\[ \tan \alpha = \frac{15 \cdot \frac{\sqrt{3}}{2}}{10 + 15 \cdot \frac{1}{2}} \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ 15 \cdot \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \]
Calculating the denominator:
\[ 10 + 15 \cdot \frac{1}{2} = 10 + 7.5 = 17.5 \]
Thus, we have:
\[ \tan \alpha = \frac{\frac{15\sqrt{3}}{2}}{17.5} \]

### Step 6: Further simplify the tangent expression
To simplify:
\[ \tan \alpha = \frac{15\sqrt{3}}{2 \cdot 17.5} = \frac{15\sqrt{3}}{35} = \frac{3\sqrt{3}}{7} \]

### Step 7: Calculate \( \alpha \)
Now, we need to find \( \alpha \):
\[ \alpha = \tan^{-1}\left(\frac{3\sqrt{3}}{7}\right) \]
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ \frac{3\sqrt{3}}{7} \approx \frac{3 \cdot 1.732}{7} \approx \frac{5.196}{7} \approx 0.742 \]
Thus,
\[ \alpha \approx \tan^{-1}(0.742) \]

### Step 8: Final calculation
Using a calculator or trigonometric tables:
\[ \alpha \approx 36.6^\circ \]

### Conclusion
The angle \( \alpha \) made by the resultant \( \vec{R} \) with vector \( \vec{P} \) is approximately \( 36.6^\circ \).