In the following, find the components of the vector `vec(PQ)` along x and y directions whose magnitude is M, and makes an angle `theta` with the x - axis :
`M = 15, theta = 30^(@)`.

Answer

To find the components of the vector \( \vec{PQ} \) along the x and y directions, we can use the following formulas based on the magnitude of the vector and the angle it makes with the x-axis.

### Given:
- Magnitude \( M = 15 \)
- Angle \( \theta = 30^\circ \)

### Step 1: Find the x-component of the vector
The x-component of the vector \( \vec{PQ} \) can be calculated using the cosine function:
\[ \text{x-component} = M \cdot \cos(\theta) \]
Substituting the values:
\[ \text{x-component} = 15 \cdot \cos(30^\circ) \]

### Step 2: Calculate \( \cos(30^\circ) \)
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Now substituting this value:
\[ \text{x-component} = 15 \cdot \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \]

### Step 3: Find the y-component of the vector
The y-component of the vector \( \vec{PQ} \) can be calculated using the sine function:
\[ \text{y-component} = M \cdot \sin(\theta) \]
Substituting the values:
\[ \text{y-component} = 15 \cdot \sin(30^\circ) \]

### Step 4: Calculate \( \sin(30^\circ) \)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
Now substituting this value:
\[ \text{y-component} = 15 \cdot \frac{1}{2} = 7.5 \]

### Final Result:
Thus, the components of the vector \( \vec{PQ} \) are:
- x-component: \( \frac{15\sqrt{3}}{2} \)
- y-component: \( 7.5 \)