ABC is a right angled triangle right angled at `BandangleA=60^(@)andAB=20cm` , then the ratio of sides BC and CA is

Options

A. `sqrt(3):1`
B. `1:sqrt(3)`
C. `sqrt(3):sqrt(2)`
D. `sqrt(3):2`

Answer

To find the ratio of sides BC and CA in triangle ABC, where angle A is 60 degrees and AB is 20 cm, we can follow these steps:

### Step 1: Identify the sides of the triangle
In triangle ABC:
- Angle A = 60 degrees
- Angle B = 90 degrees (right angle)
- Side AB = 20 cm (adjacent to angle A)
- Side BC = opposite to angle A
- Side AC = hypotenuse

### Step 2: Use the trigonometric ratio for angle A
We can use the tangent function, which relates the opposite side (BC) to the adjacent side (AB):
\[ \tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} \]
Substituting the known values:
\[ \tan(60^\circ) = \frac{BC}{20} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{BC}{20} \]

### Step 3: Solve for BC
To find BC, we multiply both sides by 20:
\[ BC = 20\sqrt{3} \]

### Step 4: Use the sine function to find AC
Next, we can use the sine function to find the hypotenuse (AC):
\[ \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} \]
Substituting the known values:
\[ \sin(60^\circ) = \frac{20\sqrt{3}}{AC} \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ \frac{\sqrt{3}}{2} = \frac{20\sqrt{3}}{AC} \]

### Step 5: Solve for AC
Cross-multiplying gives us:
\[ \sqrt{3} \cdot AC = 40\sqrt{3} \]
Dividing both sides by \(\sqrt{3}\):
\[ AC = 40 \]

### Step 6: Find the ratio of BC to AC
Now we can find the ratio of BC to AC:
\[ \text{Ratio of } BC \text{ to } AC = \frac{BC}{AC} = \frac{20\sqrt{3}}{40} = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the ratio of sides BC to AC is:
\[ \frac{\sqrt{3}}{2} \]