In a `/_\ ABC`,if `2 /_A = 3 /_B = 6 /_C`, then `/_B = ?`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations based on the given relationships.
We are given that:
- \( 2 \angle A = 3 \angle B = 6 \angle C \)

Let’s denote \( k \) as a common variable such that:
- \( 2 \angle A = k \)
- \( 3 \angle B = k \)
- \( 6 \angle C = k \)

### Step 2: Express angles A, B, and C in terms of k.
From the equations, we can express each angle in terms of \( k \):
- \( \angle A = \frac{k}{2} \)
- \( \angle B = \frac{k}{3} \)
- \( \angle C = \frac{k}{6} \)

### Step 3: Use the triangle angle sum property.
The sum of the angles in a triangle is always \( 180^\circ \). Therefore, we can write:
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the expressions we derived:
\[ \frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ \]

### Step 4: Find a common denominator and simplify.
The least common multiple of 2, 3, and 6 is 6. We can rewrite each term with a denominator of 6:
\[ \frac{3k}{6} + \frac{2k}{6} + \frac{k}{6} = 180^\circ \]
Combining the fractions:
\[ \frac{3k + 2k + k}{6} = 180^\circ \]
This simplifies to:
\[ \frac{6k}{6} = 180^\circ \]
Thus:
\[ k = 180^\circ \]

### Step 5: Calculate angle B.
Now that we have \( k \), we can find \( \angle B \):
\[ \angle B = \frac{k}{3} = \frac{180^\circ}{3} = 60^\circ \]

### Final Answer:
\[ \angle B = 60^\circ \]

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