The anlges A,B and C of a triangle ABC are in A.P and a:b=1 `sqrt(3)`.If c=4 cm then the area (in sq. cm of this triangle is :)

Options

A. `4 sqrt(3)`
B. `2/(sqrt(3)`
C. `2 sqrt(3)`
D. `4/(sqrt(3)`

Answer

To solve the problem step by step, we need to find the area of triangle ABC given that the angles A, B, and C are in arithmetic progression (A.P.), the ratio of sides a:b = 1:√3, and side c = 4 cm.

### Step 1: Determine the angles of the triangle
Since angles A, B, and C are in A.P., we can express them as:
- A = B - d
- B = B
- C = B + d

The sum of angles in a triangle is 180 degrees:
\[ A + B + C = 180 \]
Substituting the expressions for A and C:
\[ (B - d) + B + (B + d) = 180 \]
This simplifies to:
\[ 3B = 180 \implies B = 60^\circ \]
Now substituting back to find A and C:
\[ A = B - d \quad \text{and} \quad C = B + d \]
Since we know A + C = 120 degrees (180 - 60), we can find that:
\[ A + C = 120 \implies A + (120 - A) = 120 \implies A = 30^\circ, \quad C = 90^\circ \]

### Step 2: Identify the sides of the triangle using the sine rule
From the given ratio \( a:b = 1:\sqrt{3} \), we can express:
\[ a = k \quad \text{and} \quad b = k\sqrt{3} \]
for some constant \( k \).

Using the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting the known values:
\[ \frac{k}{\sin 30^\circ} = \frac{k\sqrt{3}}{\sin 60^\circ} = \frac{4}{\sin 90^\circ} \]
We know:
\[ \sin 30^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin 90^\circ = 1 \]

### Step 3: Calculate side a
Using the sine rule:
\[ \frac{k}{\frac{1}{2}} = \frac{4}{1} \implies k = 4 \times \frac{1}{2} = 2 \]
Thus, \( a = 2 \) cm.

### Step 4: Calculate side b
Now substituting \( k \) into \( b \):
\[ b = k\sqrt{3} = 2\sqrt{3} \text{ cm} \]

### Step 5: Calculate the area of the triangle
Since angle C = 90 degrees, we can use the formula for the area of a right triangle:
\[ \text{Area} = \frac{1}{2} \times a \times b \]
Substituting the values of \( a \) and \( b \):
\[ \text{Area} = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3} \text{ sq. cm} \]

### Final Answer
The area of triangle ABC is \( 2\sqrt{3} \) sq. cm.

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