Evaluate :
(i) `(cos 3A - 2 cos 4A)/(sin 3 A + 2 sin 4A)`, when `A = 15^(@)`
(ii) `(3 sin 3B + 2 cos(2B + 5^(@)))/(2 cos 3B - sin(2B - 10^(@)))`, when `B = 20^(@)`

Answer

To solve the given problems step by step, we will evaluate each part separately.

### Part (i): Evaluate \((\cos 3A - 2 \cos 4A) / (\sin 3A + 2 \sin 4A)\) when \(A = 15^\circ\)

1. **Substitute \(A = 15^\circ\)**:
- Calculate \(3A\) and \(4A\):
\[ 3A = 3 \times 15^\circ = 45^\circ \]
\[ 4A = 4 \times 15^\circ = 60^\circ \]

2. **Evaluate \(\cos 3A\) and \(\cos 4A\)**:
- \(\cos 45^\circ = \frac{1}{\sqrt{2}}\)
- \(\cos 60^\circ = \frac{1}{2}\)

3. **Substitute into the numerator**:
\[ \cos 3A - 2 \cos 4A = \frac{1}{\sqrt{2}} - 2 \times \frac{1}{2} = \frac{1}{\sqrt{2}} - 1 \]

4. **Evaluate \(\sin 3A\) and \(\sin 4A\)**:
- \(\sin 45^\circ = \frac{1}{\sqrt{2}}\)
- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)

5. **Substitute into the denominator**:
\[ \sin 3A + 2 \sin 4A = \frac{1}{\sqrt{2}} + 2 \times \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{2}} + \sqrt{3} \]

6. **Combine the results**:
\[ \frac{\cos 3A - 2 \cos 4A}{\sin 3A + 2 \sin 4A} = \frac{\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}} + \sqrt{3}} \]

7. **Multiply numerator and denominator by \(\sqrt{2}\)** to simplify:
\[ = \frac{1 - \sqrt{2}}{1 + \sqrt{6}} \]

### Final Result for Part (i):
\[ \frac{1 - \sqrt{2}}{1 + \sqrt{6}} \]

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### Part (ii): Evaluate \((3 \sin 3B + 2 \cos(2B + 5^\circ)) / (2 \cos 3B - \sin(2B - 10^\circ))\) when \(B = 20^\circ\)

1. **Substitute \(B = 20^\circ\)**:
- Calculate \(3B\) and \(2B\):
\[ 3B = 3 \times 20^\circ = 60^\circ \]
\[ 2B = 2 \times 20^\circ = 40^\circ \]

2. **Evaluate \(\sin 3B\) and \(\cos 2B + 5^\circ\)**:
- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
- \(\cos(40^\circ + 5^\circ) = \cos 45^\circ = \frac{1}{\sqrt{2}}\)

3. **Substitute into the numerator**:
\[ 3 \sin 60^\circ + 2 \cos(2B + 5^\circ) = 3 \times \frac{\sqrt{3}}{2} + 2 \times \frac{1}{\sqrt{2}} = \frac{3\sqrt{3}}{2} + \frac{2}{\sqrt{2}} \]

4. **Evaluate \(\cos 3B\) and \(\sin(2B - 10^\circ)\)**:
- \(\cos 60^\circ = \frac{1}{2}\)
- \(\sin(40^\circ - 10^\circ) = \sin 30^\circ = \frac{1}{2}\)

5. **Substitute into the denominator**:
\[ 2 \cos 60^\circ - \sin(2B - 10^\circ) = 2 \times \frac{1}{2} - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Combine the results**:
\[ \frac{3 \sin 60^\circ + 2 \cos(2B + 5^\circ)}{2 \cos 60^\circ - \sin(2B - 10^\circ)} = \frac{\frac{3\sqrt{3}}{2} + \frac{2}{\sqrt{2}}}{\frac{1}{2}} \]

7. **Multiply numerator and denominator by 2**:
\[ = 3\sqrt{3} + 2\sqrt{2} \]

### Final Result for Part (ii):
\[ 3\sqrt{3} + 2\sqrt{2} \]

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