Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining

Q: Why do we draw a horizontal line from the top of the shorter pole? Why not just connect the tops directly?

Connecting the tops directly creates a line segment, but it doesn't immediately give you a right-angled triangle with the ground. By drawing a horizontal line, you create a right-angled triangle where the angle of elevation is clearly defined. It turns a complex 2D shape into a simple, solvable triangle.

Q: Do I really need to memorize $\tan(15^\circ) = 2 - \sqrt{3}$?

In the heat of a JEE Advanced exam, speed is your best friend. While you can derive it using $\tan(45^\circ - 30^\circ)$, memorizing standard values for $15^\circ$ and $75^\circ$ saves precious minutes and reduces the chance of calculation errors.

Q: Why do we rationalize the denominator? Is it just a math convention?

It is more than convention. Rationalizing the denominator, as we did with $(2 + \sqrt{3})$, simplifies the expression into a form that matches the options provided in the exam. It transforms an awkward division into a clean, elegant product.

Question

Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

Options:

(A) $\frac{5}{2}(2+\sqrt{3})$

(B) $5(\sqrt{3}+1)$

(C) $5(2+\sqrt{3})$

(D) $10(\sqrt{3}-1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a perfectly flat, horizontal ground. In front of you, two poles rise like sentinels—one modest at $5$ meters, the other towering at $10$ meters.

The problem asks for the distance between them, given that the line connecting their tops makes an angle of $15^{\circ}$ with the horizontal.

### The Art of Construction

To solve this, we must simplify the geometry. We create a right-angled triangle by drawing a horizontal line from the top of the shorter pole to the taller pole.

This horizontal line effectively "cuts" the taller pole. The taller pole, originally $10$ meters, is now split into a lower section of $5$ meters and an upper section.

The length of this upper section, which forms the vertical side of our right-angled triangle, is:
$$10 - 5 = 5 	ext{ meters}$$

### The Trigonometric Bridge

We now have a right-angled triangle where the vertical side (opposite to the angle) is $5$ meters. The base of this triangle represents the horizontal distance between the poles, which we define as $x$.

The angle of elevation is $15^{\circ}$. We use the tangent function to relate these sides:
$$	an(	heta) = \frac{	ext{Opposite}}{	ext{Adjacent}}$$

Substituting our known values into the equation, we get:
$$	an(15^{\circ}) = \frac{5}{x}$$

### The Elegance of Algebra

To isolate $x$, we rearrange the equation:
$$x = \frac{5}{	an(15^{\circ})}$$

We utilize the known trigonometric value $	an(15^{\circ}) = 2 - \sqrt{3}$. Substituting this into our expression yields:
$$x = \frac{5}{2 - \sqrt{3}}$$

To simplify, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(2 + \sqrt{3})$:
$$x = \frac{5(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$$

Applying the difference of squares identity $(a - b)(a + b) = a^2 - b^2$ to the denominator:
$$x = \frac{5(2 + \sqrt{3})}{4 - 3}$$

The denominator simplifies to $1$, leaving us with the final result:
**$x = 5(2 + \sqrt{3}) 	ext{ meters}$**

Animated Solution for Mathematics - Trigonometry: Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

Select Answer:

Visualized Solution

Visualizing the Poles

The Unknown Distance

Connecting the Tops

Marking the Angle

The Right-Angled Triangle

Finding the Height Difference

Applying Trigonometry

Substituting the Values

Rearranging for $x$

Value of $tan (1 5^{\circ})$

Substituting $tan (1 5^{\circ})$

Rationalizing the Denominator

Simplifying the Denominator

Final Answer

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Art of Construction

The Trigonometric Bridge

The Elegance of Algebra

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