In `DeltaABC`, if `angleA:angleB:angleC=1:2:3`, then what is `BC:CA:AB?`

Options

A. `1:2:3`
B. `1:sqrt3:2`
C. `2:sqrt3:1`
D. `sqrt3:1:2`

Answer

To solve the problem of finding the ratio of the sides \( BC:CA:AB \) in triangle \( ABC \) given the angles \( \angle A:\angle B:\angle C = 1:2:3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Assign Variables to Angles**:
Let \( \angle A = x \), \( \angle B = 2x \), and \( \angle C = 3x \).

2. **Use the Triangle Angle Sum Property**:
The sum of angles in a triangle is \( 180^\circ \). Therefore, we can write the equation:
\[ x + 2x + 3x = 180^\circ \]
This simplifies to:
\[ 6x = 180^\circ \]

3. **Solve for \( x \)**:
Dividing both sides by 6 gives:
\[ x = \frac{180^\circ}{6} = 30^\circ \]
Thus, we have:
\[ \angle A = 30^\circ, \quad \angle B = 60^\circ, \quad \angle C = 90^\circ \]

4. **Apply the Sine Rule**:
According to the sine rule, the ratios of the sides opposite to the angles are proportional to the sines of those angles:
\[ \frac{BC}{\sin A} = \frac{CA}{\sin B} = \frac{AB}{\sin C} \]
Let \( k \) be the proportionality constant. Then we can express the sides as:
\[ BC = k \sin A, \quad CA = k \sin B, \quad AB = k \sin C \]

5. **Substitute the Values of Sines**:
Now substituting the values of the angles:
\[ BC = k \sin 30^\circ, \quad CA = k \sin 60^\circ, \quad AB = k \sin 90^\circ \]
Using known sine values:
\[ \sin 30^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin 90^\circ = 1 \]
Therefore:
\[ BC = k \cdot \frac{1}{2}, \quad CA = k \cdot \frac{\sqrt{3}}{2}, \quad AB = k \cdot 1 \]

6. **Form the Ratio**:
Now we can write the ratio of the sides:
\[ BC : CA : AB = \frac{k}{2} : \frac{k\sqrt{3}}{2} : k \]
Dividing through by \( k \) (assuming \( k \neq 0 \)):
\[ BC : CA : AB = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1 \]

7. **Simplify the Ratio**:
To eliminate the fractions, multiply each term by 2:
\[ BC : CA : AB = 1 : \sqrt{3} : 2 \]

### Final Answer:
Thus, the ratio \( BC:CA:AB \) is:
\[ 1 : \sqrt{3} : 2 \]