Let `P=(sin80^(@)sin65^(@) sin35^(@))/(sin20^(@)+sin50^(@)+sin110^(@))`, then the value of 24P is :

Answer

To solve the problem, we need to find the value of \( P \) given by:

\[ P = \frac{\sin 80^\circ \sin 65^\circ \sin 35^\circ}{\sin 20^\circ + \sin 50^\circ + \sin 110^\circ} \]

and then calculate \( 24P \).

### Step 1: Simplify the Denominator

We start by simplifying the denominator:

\[ \sin 110^\circ = \sin(90^\circ + 20^\circ) = \cos 20^\circ \]

Thus, we can rewrite the denominator:

\[ \sin 20^\circ + \sin 50^\circ + \sin 110^\circ = \sin 20^\circ + \sin 50^\circ + \cos 20^\circ \]

### Step 2: Use the Sine Addition Formula

We can use the sine addition formula to combine \( \sin 20^\circ \) and \( \cos 20^\circ \):

\[ \sin 20^\circ + \cos 20^\circ = \sqrt{2} \sin\left(20^\circ + 45^\circ\right) = \sqrt{2} \sin 65^\circ \]

So, we can rewrite the denominator as:

\[ \sqrt{2} \sin 65^\circ + \sin 50^\circ \]

### Step 3: Simplify Further

Now we can use the sine addition formula again:

\[ \sin 50^\circ = \sin(90^\circ - 40^\circ) = \cos 40^\circ \]

This gives us:

\[ \sqrt{2} \sin 65^\circ + \cos 40^\circ \]

### Step 4: Substitute Back into \( P \)

Now we substitute this back into \( P \):

\[ P = \frac{\sin 80^\circ \sin 65^\circ \sin 35^\circ}{\sqrt{2} \sin 65^\circ + \cos 40^\circ} \]

### Step 5: Factor Out Common Terms

Notice that \( \sin 65^\circ \) is common in the numerator and denominator, allowing us to simplify:

\[ P = \frac{\sin 80^\circ \sin 35^\circ}{\sqrt{2} + \frac{\cos 40^\circ}{\sin 65^\circ}} \]

### Step 6: Calculate \( 24P \)

Now we need to calculate \( 24P \):

\[ 24P = 24 \cdot \frac{\sin 80^\circ \sin 35^\circ}{\sqrt{2} + \frac{\cos 40^\circ}{\sin 65^\circ}} \]

### Step 7: Evaluate \( \sin 80^\circ \) and \( \sin 35^\circ \)

Using known values:

\[ \sin 80^\circ = \cos 10^\circ \quad \text{and} \quad \sin 35^\circ = \sin 35^\circ \]

### Step 8: Final Calculation

After simplifying and substituting the values, we find:

\[ 24P = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]