The vectors `vecA` has a magnitude of 5 unit `vecB` has a magnitude of 6 unit and the cross product of `vecA and vecB` has a magnitude of 15 unit. Find the angle between `vecA and vecB`.

Explanation

To find the angle between the vectors \(\vec{A}\) and \(\vec{B}\), we can use the formula for the magnitude of the cross product of two vectors:

\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(\theta) \]

Where:
- \(|\vec{A} \times \vec{B}|\) is the magnitude of the cross product,
- \(|\vec{A}|\) is the magnitude of vector \(\vec{A}\),
- \(|\vec{B}|\) is the magnitude of vector \(\vec{B}\),
- \(\theta\) is the angle between the two vectors.

### Step 1: Write down the known values
- Magnitude of \(\vec{A}\) = 5 units
- Magnitude of \(\vec{B}\) = 6 units
- Magnitude of \(\vec{A} \times \vec{B}\) = 15 units

### Step 2: Substitute the known values into the cross product formula
Using the formula:

\[ 15 = 5 \times 6 \times \sin(\theta) \]

### Step 3: Simplify the equation
Calculate \(5 \times 6\):

\[ 15 = 30 \sin(\theta) \]

### Step 4: Solve for \(\sin(\theta)\)
To isolate \(\sin(\theta)\), divide both sides by 30:

\[ \sin(\theta) = \frac{15}{30} = \frac{1}{2} \]

### Step 5: Find the angle \(\theta\)
The angle whose sine is \(\frac{1}{2}\) is:

\[ \theta = 30^\circ \]

### Final Answer
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(30^\circ\).

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