Evaluate :
`(3sin3B+2cos(2B+5^(@)))/(2cos3B-sin(2B-10^(@)))`,
when `B=20^(@)`.

Answer

To evaluate the expression

\[ \frac{3\sin(3B) + 2\cos(2B + 5^\circ)}{2\cos(3B) - \sin(2B - 10^\circ)} \]

when \( B = 20^\circ \), we will follow these steps:

### Step 1: Substitute \( B = 20^\circ \)

First, we substitute \( B \) in the expression:

\[ 3B = 3 \times 20^\circ = 60^\circ \]
\[ 2B = 2 \times 20^\circ = 40^\circ \]

Now, we can rewrite the expression as:

\[ \frac{3\sin(60^\circ) + 2\cos(40^\circ + 5^\circ)}{2\cos(60^\circ) - \sin(40^\circ - 10^\circ)} \]

### Step 2: Simplify the angles

Now, simplify the angles in the cosine and sine functions:

\[ \cos(40^\circ + 5^\circ) = \cos(45^\circ) \]
\[ \sin(40^\circ - 10^\circ) = \sin(30^\circ) \]

Now, the expression becomes:

\[ \frac{3\sin(60^\circ) + 2\cos(45^\circ)}{2\cos(60^\circ) - \sin(30^\circ)} \]

### Step 3: Calculate the trigonometric values

Using known values of trigonometric functions:

- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \cos(60^\circ) = \frac{1}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Substituting these values into the expression gives:

\[ \frac{3 \cdot \frac{\sqrt{3}}{2} + 2 \cdot \frac{1}{\sqrt{2}}}{2 \cdot \frac{1}{2} - \frac{1}{2}} \]

### Step 4: Simplify the numerator and denominator

Calculating the numerator:

\[ 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]
\[ 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

So the numerator becomes:

\[ \frac{3\sqrt{3}}{2} + \sqrt{2} \]

Calculating the denominator:

\[ 2 \cdot \frac{1}{2} = 1 \]
\[ 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 5: Final expression

Now, we have:

\[ \frac{\frac{3\sqrt{3}}{2} + \sqrt{2}}{\frac{1}{2}} = 2\left(\frac{3\sqrt{3}}{2} + \sqrt{2}\right) \]

Distributing the 2:

\[ = 3\sqrt{3} + 2\sqrt{2} \]

### Final Answer

Thus, the evaluated expression is:

\[ 3\sqrt{3} + 2\sqrt{2} \]

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