A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected

Q: Wait, why is $\triangle OAB$ necessarily an isosceles triangle?

Great observation! In right-angled $\triangle PAO$, we have $\cot 15^\circ = \frac{OA}{h}$, so $OA = h \cot 15^\circ$. Similarly, in $\triangle PBO$, we have $OB = h \cot 15^\circ$. Since $OA = OB$, the triangle $OAB$ must be isosceles. This symmetry is the key to unlocking the geometry of the base.

Q: Why can't we just use the Sine Rule on $\triangle OAB$?

You absolutely could! However, the Sine Rule would require knowing the angles of $\triangle OAB$, which are not directly given. Using the Pythagorean theorem with the altitude $OC$ is much more direct and less prone to calculation errors in this specific 3D setup.

Q: How do I handle the value of $\cot 15^\circ$?

You can derive it using the formula $\cot(45^\circ - 30^\circ) = \frac{\cot 45^\circ \cot 30^\circ + 1}{\cot 30^\circ - \cot 45^\circ}$. Substituting the values, we get $\frac{1 \cdot \sqrt{3} + 1}{\sqrt{3} - 1}$. Rationalizing this gives $2 + \sqrt{3}$. It is a very useful value to keep in your memory for competitive exams!

Question

A horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. A vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^\circ$ and $\angle PCO = 45^\circ$, where $C$ is the midpoint of $AB$. Then $(OP)^2$ is equal to

Options:

(A) $\frac{32}{\sqrt{3}}(\sqrt{3}-1)$

(B) $\frac{32}{\sqrt{3}}(2-\sqrt{3})$

(C) $\frac{16}{\sqrt{3}}(\sqrt{3}-1)$

(D) $\frac{16}{\sqrt{3}}(2-\sqrt{3})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, imagine you are standing in a beautiful, triangular park $OAB$. At the vertex $O$, a majestic lamp post $OP$ rises vertically into the sky.

This is not just a problem of numbers; it is a problem of perspective. We are bridging the gap between the 3D world of the lamp post and the 2D world of the park floor.

Let the height of the lamp post be $h = OP$. The problem provides three crucial angles: $\angle PAO = 15^\circ$, $\angle PBO = 15^\circ$, and $\angle PCO = 45^\circ$, where $C$ is the midpoint of $AB$.

### The Power of Cotangent

Look at the right-angled triangle $	riangle PAO$. Since the lamp post is vertical, $\angle POA = 90^\circ$.

Using trigonometry, we find:

$$ \cot 15^\circ = \frac{OA}{OP} = \frac{OA}{h} $$

This gives us the length $OA = h \cot 15^\circ$. By the exact same logic, in $	riangle PBO$, we find $OB = h \cot 15^\circ$.

This symmetry is beautiful, isn't it? It tells us immediately that $OA = OB$, which means $	riangle OAB$ is an isosceles triangle.

Now, turn your attention to $	riangle PCO$. Again, it is a right-angled triangle at $O$. We have:

$$ \cot 45^\circ = \frac{OC}{OP} = \frac{OC}{h} $$

Since $\cot 45^\circ = 1$, we get the elegant result $OC = h$.

### The Ground Reality

Now, let us step down from the 3D perspective to the 2D ground. We have an isosceles triangle $	riangle OAB$ where $OA = OB$.

We know that in any isosceles triangle, the median to the base is also the altitude. Since $C$ is the midpoint of $AB$, $OC$ is the median, and therefore $OC \perp AB$.

This creates a perfect right-angled triangle $	riangle OCA$ on the ground. The length $AC$ is simply half of $AB$, so $AC = \frac{16}{2} = 8$.

Now, we apply the Pythagorean theorem:

$$ OA^2 = OC^2 + AC^2 $$

Substituting our expressions, we get:

$$ (h \cot 15^\circ)^2 = h^2 + 8^2 $$

### The Algebraic Finale

We know $\cot 15^\circ = 2 + \sqrt{3}$. Substituting this, the equation becomes:

$$ h^2(2 + \sqrt{3})^2 = h^2 + 64 $$

Expanding the square, $(2 + \sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$. So, $h^2(7 + 4\sqrt{3}) = h^2 + 64$.

Rearranging to isolate $h^2$, we get:

$$ h^2(7 + 4\sqrt{3} - 1) = 64 $$

This simplifies to $h^2(6 + 4\sqrt{3}) = 64$. Dividing by 2, we have $h^2(3 + 2\sqrt{3}) = 32$, or:

$$ h^2 = \frac{32}{3 + 2\sqrt{3}} $$

To match the options, we rationalize the denominator by multiplying by $\frac{2\sqrt{3} - 3}{2\sqrt{3} - 3}$:

$$ h^2 = \frac{32(2\sqrt{3} - 3)}{12 - 9} = \frac{32(2\sqrt{3} - 3)}{3} $$

This can be rewritten as $\frac{32}{\sqrt{3}}(2 - \sqrt{3})$. And there it is! The complexity dissolves into a simple, elegant expression.

Remember, in JEE Advanced, the most complex 3D problems are often just a collection of simple 2D triangles waiting to be discovered. Keep practicing, and keep visualizing!

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Analyzing Right $△ P A O$

Analyzing Right $△ P C O$

Geometry of the Base Triangle

Applying Pythagoras Theorem

Evaluating $cot 1 5^{\circ}$

Expanding the Square

Isolating $h^{2}$

Simplifying the Fraction

Rationalizing the Denominator

Final Rationalization Step

Final Answer

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Power of Cotangent

The Ground Reality

The Algebraic Finale

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