In a `Delta ABC`, if `2 /_A = 3/_B = 6 /_C`, value of `/_B` is

Options

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

Answer

To solve the problem, we start with the given relationships between the angles of triangle ABC:

1. \( 2 \cdot \angle A = 3 \cdot \angle B = 6 \cdot \angle C \)

Let's denote a common variable \( x \) such that:

- \( 2 \cdot \angle A = x \)
- \( 3 \cdot \angle B = x \)
- \( 6 \cdot \angle C = x \)

From these equations, we can express each angle in terms of \( x \):

2. \( \angle A = \frac{x}{2} \)
3. \( \angle B = \frac{x}{3} \)
4. \( \angle C = \frac{x}{6} \)

Next, we know that the sum of the angles in a triangle is always \( 180^\circ \):

5. \( \angle A + \angle B + \angle C = 180^\circ \)

Substituting the expressions for \( \angle A \), \( \angle B \), and \( \angle C \) into this equation:

6. \( \frac{x}{2} + \frac{x}{3} + \frac{x}{6} = 180^\circ \)

To solve this equation, we need a common denominator. The least common multiple of 2, 3, and 6 is 6. Therefore, we can rewrite each term:

7. \( \frac{3x}{6} + \frac{2x}{6} + \frac{x}{6} = 180^\circ \)

Now, combine the fractions:

8. \( \frac{3x + 2x + x}{6} = 180^\circ \)

This simplifies to:

9. \( \frac{6x}{6} = 180^\circ \)

Thus, we find:

10. \( x = 180^\circ \)

Now we can find each angle:

11. \( \angle A = \frac{x}{2} = \frac{180^\circ}{2} = 90^\circ \)
12. \( \angle B = \frac{x}{3} = \frac{180^\circ}{3} = 60^\circ \)
13. \( \angle C = \frac{x}{6} = \frac{180^\circ}{6} = 30^\circ \)

Finally, the value of \( \angle B \) is:

14. \( \angle B = 60^\circ \)

### Final Answer:
The value of \( \angle B \) is \( 60^\circ \).