In `DeltaABC, a = 3, b = 4` and `c = 5`, then value of `sinA + sin2B + sin3C` is

Options

A. `(24)/(25)`
B. `(14)/(25)`
C. `(64)/(25)`
D. none

Answer

To solve the problem, we need to find the value of \( \sin A + \sin 2B + \sin 3C \) for the triangle \( \Delta ABC \) where \( a = 3 \), \( b = 4 \), and \( c = 5 \).

### Step 1: Find \( \sin A \)

Using the sine rule, we can find \( \sin A \):
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]
Since \( c = 5 \) and \( a = 3 \), we can write:
\[ \sin A = \frac{a \cdot \sin C}{c} \]
First, we need to find \( \sin C \). In a right triangle, we can use the Pythagorean theorem to find angles. Since \( a^2 + b^2 = c^2 \) holds true, \( \Delta ABC \) is a right triangle with \( C = 90^\circ \).

Thus, \( \sin C = 1 \):
\[ \sin A = \frac{3 \cdot 1}{5} = \frac{3}{5} \]

### Step 2: Find \( \sin B \)

Using the sine rule again:
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting the known values:
\[ \sin B = \frac{b \cdot \sin C}{c} = \frac{4 \cdot 1}{5} = \frac{4}{5} \]

### Step 3: Find \( \sin 2B \)

Using the double angle formula:
\[ \sin 2B = 2 \sin B \cos B \]
We need to find \( \cos B \):
\[ \cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left( \frac{4}{5} \right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]
Now substituting \( \sin B \) and \( \cos B \):
\[ \sin 2B = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25} \]

### Step 4: Find \( \sin 3C \)

Since \( C = 90^\circ \):
\[ \sin 3C = \sin 270^\circ = -1 \]

### Step 5: Combine the results

Now we can combine all the results:
\[ \sin A + \sin 2B + \sin 3C = \frac{3}{5} + \frac{24}{25} - 1 \]
To combine these, we need a common denominator:
\[ \frac{3}{5} = \frac{15}{25} \]
Thus:
\[ \sin A + \sin 2B + \sin 3C = \frac{15}{25} + \frac{24}{25} - \frac{25}{25} = \frac{15 + 24 - 25}{25} = \frac{14}{25} \]

### Final Answer

The value of \( \sin A + \sin 2B + \sin 3C \) is:
\[ \frac{14}{25} \]

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