Animated Solution for Mathematics - Trigonometry: If the angles of a triangle are $30^{\circ }$ and $45^{\circ }$ and the included side is $(\sqrt{3}+1)$ cms, then the area of the triangle is ..................

My dear student, welcome to a beautiful exploration of geometry. When you look at a triangle with angles $30^{\circ }$ and $45^{\circ }$ and an included side of $(\sqrt{3}+1)$, do not just see numbers. See a structure waiting to be unlocked.

In JEE Advanced, the most complex problems often yield to the simplest geometric constructions. Today, we are going to peel back the layers of this triangle using the power of the altitude.

The Altitude Strategy

We are given a triangle $ABC$ where the side $c=AB=\sqrt{3}+1$. We know $\angle A=30^{\circ }$ and $\angle B=45^{\circ }$.

The area of any triangle is fundamentally defined as:

We have the base, but the height is hidden. To reveal it, we drop a perpendicular from vertex $C$ to the base $AB$, meeting it at point $D$. Let the length of this altitude $CD$ be $h$.

This single line is the key; it transforms our non-right triangle into two right-angled triangles: $\Delta ACD$ and $\Delta BCD$.

The Trigonometric Bridge

Now, let us focus on $\Delta ACD$. We have a base angle of $30^{\circ }$ and an opposite side $h$. The adjacent side is $AD$.

Using the cotangent ratio:

Since $\cot (30^{\circ })=\sqrt{3}$, we find that $AD=h\sqrt{3}$.

Next, we turn to $\Delta BCD$. The base angle is $45^{\circ }$, and the opposite side is $h$. The adjacent side is $DB$.

Again, using the cotangent ratio:

Since $\cot (45^{\circ })=1$, we get $DB=h$. This is the beauty of the $45^{\circ }$ angle—it creates an isosceles right triangle where the height and the base segment are identical.

The Algebraic Symphony

We know that the total base $AB$ is the sum of its segments: $AB=AD+DB$. Substituting our expressions, we get:

Now, factor out the $h$:

Look at the elegance of this moment! The term $(\sqrt{3}+1)$ appears on both sides. Dividing both sides by $(\sqrt{3}+1)$, we find that $h=1$. The height is exactly $1$ unit.

The Final Calculation

With the height $h=1$ and the base $AB=\sqrt{3}+1$, the area is simply:

Thus, the area is $\frac{\sqrt{3}+1}{2}$ square units.

You see, by constructing the altitude, we turned a daunting problem into a simple algebraic equation. Keep this strategy in your toolkit—whenever you see a triangle, look for the hidden height. You have done excellent work today.