If the sides of a triangle are in the ratio 1 : `sqrt3` : 2, then the angles of the triangle are in the ratio

Options

A. `1:3:5`
B. `2:3:1`
C. `3:2:1`
D. `1:2:3`

Explanation

To solve the problem of finding the ratio of the angles of a triangle when the sides are in the ratio \(1 : \sqrt{3} : 2\), we can follow these steps:

### Step 1: Assign Variables to the Sides
Let the sides of the triangle be represented as:
- \(a = k\)
- \(b = \sqrt{3}k\)
- \(c = 2k\)

Here, \(k\) is a proportionality constant.

### Step 2: Apply the Pythagorean Theorem
We can check if the triangle is a right triangle by using the Pythagorean theorem. According to this theorem, for a triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the longest side), the following should hold:
\[ a^2 + b^2 = c^2 \]

Substituting the values we assigned:
\[ (k)^2 + (\sqrt{3}k)^2 = (2k)^2 \]
This simplifies to:
\[ k^2 + 3k^2 = 4k^2 \]
\[ 4k^2 = 4k^2 \]
This confirms that the triangle is a right triangle with \(c\) as the hypotenuse.

### Step 3: Identify the Angles
Since we have established that the triangle is a right triangle, we know that one angle (let's call it \(C\)) is \(90^\circ\).

### Step 4: Use the Sine Rule
Using the sine rule, we can find the other angles \(A\) and \(B\):
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting the known values:
\[ \frac{k}{\sin A} = \frac{\sqrt{3}k}{\sin B} = \frac{2k}{\sin 90^\circ} \]
Since \(\sin 90^\circ = 1\), we have:
\[ \frac{k}{\sin A} = \frac{2k}{1} \implies \sin A = \frac{k}{2k} = \frac{1}{2} \]
Thus, \(A = 30^\circ\).

For angle \(B\):
\[ \frac{\sqrt{3}k}{\sin B} = \frac{2k}{1} \implies \sin B = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \]
Thus, \(B = 60^\circ\).

### Step 5: Write the Angles in Ratio
Now we can summarize the angles:
- \(A = 30^\circ\)
- \(B = 60^\circ\)
- \(C = 90^\circ\)

The ratio of the angles \(A : B : C\) is:
\[ 30 : 60 : 90 \]
This can be simplified to:
\[ 1 : 2 : 3 \]

### Final Answer
The angles of the triangle are in the ratio \(1 : 2 : 3\).

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