Animated Solution for Mathematics - Trigonometry: A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $\alpha$ and $\beta$ at $A$ and $B$ respectively. $AB$ subtends an angle $\gamma$ at $Q$. Find the height of the pole.

Welcome, future engineer. Today, we are not just solving a trigonometry problem; we are learning to navigate the third dimension. Many students stumble when they see a problem involving a pole, angles of elevation, and ground distances because they try to force everything into a single 2D sketch.

But the secret to JEE Advanced physics and math is the ability to decompose complex 3D reality into manageable 2D slices. Let’s embark on this journey together.

Imagine you are standing on a flat, infinite plain. In front of you stands a vertical pole, $PQ$, with height $h$. This pole is our anchor.

We have two observers, $A$ and $B$, standing on the ground. When they look up at the top of the pole ($P$), they see it at angles of elevation $\alpha$ and $\beta$, respectively.

Here is the first trap: do not try to draw the triangle $PAB$ immediately. It is a distraction. Instead, focus on the two right-angled triangles formed by the pole and the ground: $\Delta PQA$ and $\Delta PQB$.

Because the pole is vertical, the angle at the base $Q$ is $90^{\circ }$ for both triangles. In $\Delta PQA$, we have the relationship:

Rearranging this, we find the distance from the base of the pole to point $A$ is $AQ=h\cot \alpha$. Similarly, for $\Delta PQB$, we find $BQ=h\cot \beta$.

Why did we use $\cot$? Because it keeps our unknown $h$ in the numerator. In the heat of an exam, keeping your variables clean is half the battle. We have now successfully translated the vertical information into the horizontal plane.

Now, let's shift our perspective. Forget the pole for a moment. Look down at the ground. We have a triangle $AQB$ lying flat on the horizontal surface.

We know the length of the side $AB$ is $d$. We know the lengths of the other two sides, $AQ$ and $BQ$, in terms of our unknown height $h$. And most importantly, we are given that the line segment $AB$ subtends an angle $\gamma$ at the base $Q$.

This is the moment where the geometry clicks. We have a triangle where we know two sides and the included angle. This is the classic setup for the Law of Cosines.

If you ever feel lost in a geometry problem, look for the 'Side-Angle-Side' (SAS) configuration. It is the bridge that connects the unknown to the known.

We apply the Law of Cosines to $\Delta AQB$:

Now, we substitute our expressions for $AQ$ and $BQ$ into this equation. This is where precision matters. Do not rush. Substitute carefully:

When we expand this, we get:

Look at the right-hand side. Every single term contains an $h^2$. This is not a coincidence; it is the mathematical structure of the problem revealing itself to you. We can factor out $h^2$:

We are almost there. To isolate $h^2$, we divide $d^2$ by the entire trigonometric expression in the parentheses. Finally, taking the square root of both sides gives us the height of the pole:

Take a moment to look at this result. It is elegant, symmetric, and perfectly derived from the geometry of the problem. You didn't just memorize a formula; you built it from the ground up.

You visualized the 3D space, decomposed it into 2D triangles, applied the Law of Cosines, and algebraically solved for the unknown. This is the essence of the JEE Advanced mindset. Keep this clarity, keep this patience, and you will conquer any problem that comes your way.