If in a triangle `ABC, 3 sin A = 6 sin B=2sqrt3sin C`, then the angle A is

Options

A. `0^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we start with the given equation:

**Step 1: Set up the equation**
Given:
\[ 3 \sin A = 6 \sin B = 2\sqrt{3} \sin C \]

**Step 2: Simplify the equation**
Divide the entire equation by 3:
\[ \sin A = 2 \sin B = \frac{2\sqrt{3}}{3} \sin C \]

**Step 3: Express sin values in terms of sides**
Using the sine rule, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
Where \( R \) is the circumradius of the triangle.

From the sine rule, we can express:
\[ \sin A = \frac{a}{2R}, \quad \sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R} \]

**Step 4: Substitute the sine values**
Substituting these into our simplified equation:
\[ \frac{a}{2R} = 2 \cdot \frac{b}{2R} = \frac{2\sqrt{3}}{3} \cdot \frac{c}{2R} \]

This simplifies to:
\[ a = 2b \quad \text{and} \quad a = \frac{2\sqrt{3}}{3}c \]

**Step 5: Relate the sides**
From \( a = 2b \), we can express \( b \) in terms of \( a \):
\[ b = \frac{a}{2} \]

From \( a = \frac{2\sqrt{3}}{3}c \), we can express \( c \) in terms of \( a \):
\[ c = \frac{3a}{2\sqrt{3}} \]

**Step 6: Use the cosine rule**
Now, we can use the cosine rule to find \( \cos A \):
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substituting the values of \( b \) and \( c \):
\[ b = \frac{a}{2}, \quad c = \frac{3a}{2\sqrt{3}} \]

**Step 7: Calculate \( b^2 \) and \( c^2 \)**
Calculating \( b^2 \) and \( c^2 \):
\[ b^2 = \left(\frac{a}{2}\right)^2 = \frac{a^2}{4} \]
\[ c^2 = \left(\frac{3a}{2\sqrt{3}}\right)^2 = \frac{9a^2}{12} = \frac{3a^2}{4} \]

**Step 8: Substitute into cosine rule**
Now substituting into the cosine rule:
\[ \cos A = \frac{\frac{a^2}{4} + \frac{3a^2}{4} - a^2}{2 \cdot \frac{a}{2} \cdot \frac{3a}{2\sqrt{3}}} \]
\[ = \frac{a^2 - a^2}{\frac{3a^2}{2\sqrt{3}}} = 0 \]

**Step 9: Solve for angle A**
Since \( \cos A = 0 \), it follows that:
\[ A = 90^\circ \]

Thus, the angle \( A \) is \( 90^\circ \).

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