Animated Solution for Mathematics - Trigonometry: The angles 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 :

We begin with the most elegant clue: the angles $A$, $B$, and $C$ are in an Arithmetic Progression. In the language of mathematics, this means there is a common difference between consecutive terms.

If $A,B,$ and $C$ are in A.P., then the middle term $B$ must be the average of the other two. Mathematically, we write this as:

We know the fundamental law of all triangles: the sum of their interior angles is always $180^{\circ }$. So, $A+B+C=180^{\circ }$.

By substituting our A.P. property ($A+C=2B$) into this sum, we get:

This simplifies to $3B=180^{\circ }$. Just like that, we have cracked the first code: $B=60^{\circ }$.

Now that we have anchored our triangle with $B=60^{\circ }$, we turn our attention to the sides. We are given the ratio $a:b=1:\sqrt{3}$.

This is a classic invitation to use the Sine Rule, which states:

By rearranging this, we find that $\frac{a}{b}=\frac{\sin A}{\sin B}$. Substituting our known values, we have:

Since $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, our equation becomes:

This is a beautiful moment of cancellation! If $\sin A=\frac{1}{2}$, then $A=30^{\circ }$.

With $A=30^{\circ }$ and $B=60^{\circ }$, the third angle $C$ is forced into existence:

The triangle is not just any triangle; it is a right-angled triangle! We know the hypotenuse $c=4$.

Using the definitions of sine and cosine, we find the legs:

We have arrived at the final step. The area of a right-angled triangle is simply $\frac{1}{2}\times \text{base}\times \text{height}$.

With our base $a=2$ and height $b=2\sqrt{3}$, the area is:

We have navigated the arithmetic progression, bridged the gap with the Sine Rule, and discovered the right-angled heart of the triangle. The final area is $2\sqrt{3}$ sq. cm.