Animated Solution for Mathematics - Trigonometry: In a triangle PQR, let $\angle PQR=30^{\circ }$ and the sides PQ and QR have lengths $10\sqrt{3}$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

Unlocking Triangle $PQR$

Welcome, fellow explorer of mathematics! Today, we are going to dissect a beautiful geometry problem. It is not just about finding numbers; it is about uncovering the hidden architecture of a triangle.

Imagine you are standing on a plane, and you have two rods of lengths $10\sqrt{3}$ and $10$, joined at an angle of $30^{\circ }$. This is our triangle $PQR$. Let's embark on this journey to understand its properties.

The Bridge to the Third Side

We start with two sides and an included angle. This is the perfect setup for the Cosine Rule.

Think of the Cosine Rule as a generalized version of the Pythagorean theorem that works for any triangle, not just right-angled ones. It acts as a bridge, connecting the two sides we know to the one we don't. We write it as:

Substituting our values, we have $PQ=10\sqrt{3}$ and $QR=10$. The calculation unfolds:

Since $\cos (30^{\circ })=\frac{\sqrt{3}}{2}$, the expression becomes $300+100-200\sqrt{3}\cdot \frac{\sqrt{3}}{2}$. The $\sqrt{3}\cdot \sqrt{3}$ becomes $3$, and the $2$ in the denominator cancels with the $200$, leaving us with $400-300=100$.

Thus, $PR=10$. We have found our missing side!

The Isosceles Revelation

Look at what we have achieved. We found $PR=10$, and we were given $QR=10$. This is a moment of mathematical beauty—the triangle is isosceles!

Because $PR=QR$, the angles opposite these sides must be equal. This means $\angle P=\angle Q=30^{\circ }$.

Since the sum of angles in a triangle is $180^{\circ }$, the third angle, $\angle R$, must be $180^{\circ }-(30^{\circ }+30^{\circ })=120^{\circ }$. This simple realization saves us so much time and confirms that option (A) is false.

Area and the Inradius

Now, let's calculate the area, $\Delta$. The formula $\Delta =\frac{1}{2}ab\sin C$ is our best friend here. Using sides $PQ$ and $QR$ and the included angle $Q$:

This confirms option (B) is true. Next, we calculate the inradius $r$. We use $r=\frac{\Delta }{s}$, where $s$ is the semi-perimeter.

The perimeter is $10\sqrt{3}+10+10=20+10\sqrt{3}$, so $s=10+5\sqrt{3}$. Plugging these in:

By multiplying the numerator and denominator by the conjugate $(2-\sqrt{3})$, we get $r=5\sqrt{3}(2-\sqrt{3})=10\sqrt{3}-15$. This confirms option (C) is true.

The Grand Finale

Finally, we look at the circumcircle. The circumradius $R$ is given by $R=\frac{a}{2\sin A}$. Using side $QR$ and angle $P$:

The area of the circumcircle is $\pi R^2=\pi (10{)}^2=100\pi$. Option (D) is also true!

We have successfully navigated the geometry of this triangle, proving that B, C, and D are the correct statements. Keep practicing, and you will find that geometry is not just about solving problems—it is about seeing the hidden order in the world.