In a `Delta` ABC, if `2 /_A = 3 /_B = 6 /_C` then calculate `/_A, /_B` and `/_C`.

Options

A. `/_A= 40^(@), /_B= 60^(@)` and `/_C= 30^(@)`
B. `/_A= 60^(@), /_B= 60^(@)` and `/_C= 30^(@)`
C. `/_A= 90^(@), /_B= 60^(@)` and `/_C= 30^(@)`
D. `/_A= 90^(@), /_B= 90^(@)` and `/_C= 30^(@)`

Answer

To solve the problem where \(2 \angle A = 3 \angle B = 6 \angle C\) in triangle ABC, we can follow these steps:

### Step 1: Set up the equations
We can express the angles in terms of a common variable \(K\):
- Let \(2 \angle A = K\)
- Let \(3 \angle B = K\)
- Let \(6 \angle C = K\)

### Step 2: Express each angle in terms of \(K\)
From the equations above, we can express each angle as follows:
- \( \angle A = \frac{K}{2} \)
- \( \angle B = \frac{K}{3} \)
- \( \angle C = \frac{K}{6} \)

### 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 for the angles:
\[ \frac{K}{2} + \frac{K}{3} + \frac{K}{6} = 180^\circ \]

### Step 4: Find a common denominator
The least common multiple of the denominators (2, 3, and 6) is 6. We will convert each term to have a denominator of 6:
\[ \frac{3K}{6} + \frac{2K}{6} + \frac{K}{6} = 180^\circ \]

### Step 5: Combine the fractions
Now, we can combine the fractions:
\[ \frac{3K + 2K + K}{6} = 180^\circ \]
This simplifies to:
\[ \frac{6K}{6} = 180^\circ \]
Thus, we have:
\[ K = 180^\circ \]

### Step 6: Calculate each angle
Now that we have \(K\), we can find each angle:
- \( \angle A = \frac{K}{2} = \frac{180^\circ}{2} = 90^\circ \)
- \( \angle B = \frac{K}{3} = \frac{180^\circ}{3} = 60^\circ \)
- \( \angle C = \frac{K}{6} = \frac{180^\circ}{6} = 30^\circ \)

### Final Answer
Therefore, the angles of triangle ABC are:
- \( \angle A = 90^\circ \)
- \( \angle B = 60^\circ \)
- \( \angle C = 30^\circ \)

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