If the angle 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. `4sqrt(3)`
C. `2sqrt(3)`
D. `(2)/(sqrt(3))`

Answer

To solve the problem step by step, we will follow the given information and apply the relevant mathematical concepts.

### Step 1: Understand the Given Information
We are given that the angles A, B, and C of triangle ABC are in Arithmetic Progression (A.P.) and the ratio of the sides a:b = 1:√3. We also know that side c = 4 cm.

### Step 2: Express the Angles in Terms of A
Since A, B, and C are in A.P., we can express them as:
- A = A
- B = A + d (where d is the common difference)
- C = A + 2d

The sum of angles in a triangle is 180 degrees (or π radians):
\[ A + B + C = 180^\circ \]
Substituting the expressions for B and C:
\[ A + (A + d) + (A + 2d) = 180^\circ \]
This simplifies to:
\[ 3A + 3d = 180^\circ \]
Dividing through by 3 gives:
\[ A + d = 60^\circ \]
Thus, we can express d as:
\[ d = 60^\circ - A \]

### Step 3: Find the Angles
Now substituting d back into the expressions for B and C:
- B = A + (60^\circ - A) = 60^\circ
- C = A + 2(60^\circ - A) = 120^\circ - A

### Step 4: Use the Sine Rule
Using the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Given the ratio a:b = 1:√3, we can write:
\[ \frac{a}{b} = \frac{1}{\sqrt{3}} \]
This implies:
\[ \frac{\sin A}{\sin B} = \frac{1}{\sqrt{3}} \]

### Step 5: Substitute Known Angles
We know B = 60°:
\[ \frac{\sin A}{\sin 60^\circ} = \frac{1}{\sqrt{3}} \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ \frac{\sin A}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]
Cross-multiplying gives:
\[ \sin A = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2} \]
Thus, \( A = 30^\circ \).

### Step 6: Find Remaining Angles
Now we can find B and C:
- B = 60°
- C = 120° - 30° = 90°

### Step 7: Identify the Triangle Type
Since angle C = 90°, triangle ABC is a right triangle.

### Step 8: Calculate the Area of the Triangle
The area of a right triangle is given by:
\[ \text{Area} = \frac{1}{2} \times a \times b \]
We need to find sides a and b. Using the sine rule:
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]
Substituting known values:
\[ \frac{a}{\frac{1}{2}} = \frac{4}{1} \]
Thus:
\[ a = 4 \cdot \frac{1}{2} = 2 \text{ cm} \]

Now, using the ratio \( a:b = 1:\sqrt{3} \):
\[ \frac{2}{b} = \frac{1}{\sqrt{3}} \]
Cross-multiplying gives:
\[ b = 2\sqrt{3} \text{ cm} \]

### Step 9: Final Area Calculation
Now substituting a and b into the area formula:
\[ \text{Area} = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3} \text{ sq. cm} \]

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

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