In a triangle ABC if `tanA: tan B : tan C = 3:4:5` then the value of sin A sin B sin C is

Options

A. `( 2)/( sqrt( 5))`
B. `( 2 sqrt( 5))/( 3)`
C. `( 2sqrt( 5))/( 9)`
D. `( 2)/( 3 sqrt( 5))`

Answer

To solve the problem, we start with the given ratio of the tangents of angles A, B, and C in triangle ABC:

\[ \tan A : \tan B : \tan C = 3 : 4 : 5 \]

### Step 1: Assign Variables
Let:
\[ \tan A = 3k, \quad \tan B = 4k, \quad \tan C = 5k \]

### Step 2: Use the Identity for Tangents in a Triangle
In any triangle, the sum of the angles is \(A + B + C = \pi\). Using the tangent addition formula, we can express the sum of the tangents:
\[ \tan A + \tan B + \tan C = \tan(A + B + C) = \tan(\pi) = 0 \]
Thus, we have:
\[ 3k + 4k + 5k = 0 \]
This does not directly help us, so we will use the product of tangents instead.

### Step 3: Find the Value of \(k\)
Using the identity for the product of tangents in a triangle:
\[ \tan A \tan B \tan C = \tan A + \tan B + \tan C \]
Substituting our expressions:
\[ (3k)(4k)(5k) = 3k + 4k + 5k \]
This simplifies to:
\[ 60k^3 = 12k \]
Dividing both sides by \(k\) (assuming \(k \neq 0\)):
\[ 60k^2 = 12 \]
Thus, we find:
\[ k^2 = \frac{12}{60} = \frac{1}{5} \quad \Rightarrow \quad k = \frac{1}{\sqrt{5}} \]

### Step 4: Calculate \(\tan A\), \(\tan B\), and \(\tan C\)
Now substituting \(k\) back into our expressions for \(\tan A\), \(\tan B\), and \(\tan C\):
\[ \tan A = 3k = \frac{3}{\sqrt{5}}, \quad \tan B = 4k = \frac{4}{\sqrt{5}}, \quad \tan C = 5k = \frac{5}{\sqrt{5}} \]

### Step 5: Find Sine Values
Using the relationship \(\sin A = \frac{\tan A}{\sqrt{1+\tan^2 A}}\):
\[ \sin A = \frac{\frac{3}{\sqrt{5}}}{\sqrt{1 + \left(\frac{3}{\sqrt{5}}\right)^2}} = \frac{\frac{3}{\sqrt{5}}}{\sqrt{1 + \frac{9}{5}}} = \frac{\frac{3}{\sqrt{5}}}{\sqrt{\frac{14}{5}}} = \frac{3}{\sqrt{14}} \]

Similarly, for \(\sin B\) and \(\sin C\):
\[ \sin B = \frac{\frac{4}{\sqrt{5}}}{\sqrt{1 + \left(\frac{4}{\sqrt{5}}\right)^2}} = \frac{4}{\sqrt{21}} \]
\[ \sin C = \frac{\frac{5}{\sqrt{5}}}{\sqrt{1 + \left(\frac{5}{\sqrt{5}}\right)^2}} = \frac{5}{\sqrt{30}} \]

### Step 6: Calculate \( \sin A \sin B \sin C \)
Now we calculate the product:
\[ \sin A \sin B \sin C = \left(\frac{3}{\sqrt{14}}\right) \left(\frac{4}{\sqrt{21}}\right) \left(\frac{5}{\sqrt{30}}\right) \]
Calculating the numerator:
\[ 3 \times 4 \times 5 = 60 \]
Calculating the denominator:
\[ \sqrt{14} \times \sqrt{21} \times \sqrt{30} = \sqrt{14 \times 21 \times 30} \]
Calculating \(14 \times 21 = 294\) and \(294 \times 30 = 8820\):
\[ \sin A \sin B \sin C = \frac{60}{\sqrt{8820}} \]

### Step 7: Simplify
Now we simplify \(\sqrt{8820}\):
\[ 8820 = 2 \times 3^2 \times 7 \times 5 \times 14 \]
Thus:
\[ \sqrt{8820} = 2 \times 3 \times \sqrt{70} = 6\sqrt{70} \]
Finally:
\[ \sin A \sin B \sin C = \frac{60}{6\sqrt{70}} = \frac{10}{\sqrt{70}} = \frac{10\sqrt{70}}{70} = \frac{\sqrt{70}}{7} \]

### Final Answer
Thus, the value of \(\sin A \sin B \sin C\) is:
\[ \frac{2\sqrt{5}}{7} \]