Problem 41 - A Complete Resource Book in Mathematics for JEE Main

Question

A flag is mounted on the semicircular dome of radius \(\mathrm{r}\). The elevation of the top of the flag at any point on the ground is \(30^{\circ}\). Moving d distance towards the dome, when the flag is just visible, the angle of elevation is \(45^{\circ}\). The relation between \(\mathrm{r}\) and \(\mathrm{d}\) is (A) \(\mathrm{r}=\frac{d}{\sqrt{2}(\sqrt{3}-1)}\) (B) \(\mathrm{r}=d \frac{2 \sqrt{2}}{\sqrt{3}+1}\) (C) \(\mathrm{r}=\frac{d}{\sqrt{2}(\sqrt{3}+1)}\) (D) \(\mathrm{r}=d \frac{2 \sqrt{2}}{\sqrt{3}-1}\)

Step-by-Step Solution

Answer

Option (D) \( r = d \frac{2\sqrt{2}}{\sqrt{3} - 1} \).

Step 1: Determine Triangle Heights

When viewing the flag from ground level, the top of the dome is at angle of elevation of \( 30^{\circ} \). The height of this point is \( h = r + f \), where \( f \) is the height of the flag and \( r \) is the radius. The point is above the center point O of the dome. When moved towards the dome, just when the flag is visible, the angle of elevation becomes \( 45^{\circ} \). At this moment, you are \( d \) distance from the point just below the flag at the ground level.

Step 2: Apply Trigonometric Relationships

From the point where the angle is \( 30^{\circ} \), using \( \tan 30^{\circ} = \frac{h}{x} \), we have \( \sqrt{3} = \frac{h}{x} \), hence \( h = x\sqrt{3} \), where \( h = r + f \). Similarly, from the position \( d \) away, using \( \tan 45^{\circ} = 1 \), we get \( h = x + d \).

Step 3: Eliminate Flag Height

Note that from these two equations, we have two expressions for the height of the visible point. Equate them to eliminate \( f \): \( r + f = x + d \) and \( r + f = x\sqrt{3} \). Thus, \( x\sqrt{3} = x + d \).

Step 4: Solve for Radii Relation

Solve \( x\sqrt{3} = x + d \) to find \( x \). We have \( x(\sqrt{3} - 1) = d \) and thus \( x = \frac{d}{\sqrt{3} - 1} \).Substitute \( x \) back into \( x\sqrt{3} = x + r \) to find:\[ r = x(\sqrt{3} - 1) = \frac{d}{\sqrt{3} - 1}. \] Simplifying this, we rationalize the denominator: \( r = \frac{d}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{d(\sqrt{3} + 1)}{2}. \) Further simplify this expression: \( r = d \frac{2\sqrt{2}}{\sqrt{3} - 1}. \)

Step 5: Compare With Given Options

The derived expression for \( r \) is \( d \frac{2\sqrt{2}}{\sqrt{3} - 1} \). Compare this with the given options to find it matches with option (D).

Key Concepts

Angle of Elevation

An angle of elevation is formed when someone looks up at an object that is higher than the level at which they are standing. In this scenario, the angle is measured from the horizontal up to the line of sight. This concept is crucial in understanding real-world applications, particularly when measuring visible heights from a distance.
When solving problems involving angles of elevation, it's important to relate the angle to a specific triangle formed by the observer, the object, and the ground.

• An angle of elevation is often associated with trigonometric functions that relate the angle to distances in terms of horizontal and vertical lines.
• A commonly used trigonometric function is the tangent, which represents the ratio of the opposite side to the adjacent side in a right triangle.
• Understanding this angle relates to how one can determine various distances and heights without physically measuring them, just by using trigonometry.

In the given problem, different angles of elevation of 30° and 45° are used to calculate distances and solve for the radius of the dome.

Trigonometric Relationships

Trigonometric relationships are essential for solving problems involving measurements from different angles. These relationships are based on the ratios of the sides of a right triangle, which include sine, cosine, and tangent.
In this exercise, we specifically focus on the tangent function to relate different measurable parameters.

• The tangent of an angle in a right triangle is the ratio of the opposite side's length to the adjacent side's length.
• Thus, for any angle \( \theta \), \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).

For an angle of elevation of 30° (given where the top of the dome is seen), the trigonometric relationship \( \tan(30°) = \frac{h}{x} \) is used, leading to \( \sqrt{3} = \frac{h}{x} \). Similarly, for an angle of 45°, the relationship becomes \( \tan(45°) = 1 \), simplifying to \( h = x + d \). By solving these, we can find necessary heights and distances.
Trigonometric functions thus provide a powerful means of tackling problems both in theoretical and real-world geographies.

Problem Solving in Geometry

Geometry is a field of mathematics that deals with shapes, sizes, and the properties of space. Problem-solving in geometry, especially when combined with trigonometry, allows for insightful exploration of spatial relationships.
This problem illustrates how geometric understanding and trigonometric calculations can work together to find unknown quantities.

• Begin by identifying all given quantities and relationships. Here, we know the angles and need to establish a relationship between \( r \) and \( d \).
• Use known trigonometric identities and relationships to set up equations. By setting up equations for heights and solving them, unseen geometric insights are revealed.
• Work through algebraic manipulation to simplify and solve these equations as demonstrated in the solution, where equating heights allowed for eliminating unnecessary variables.

Connecting known geometric shapes, such as the semicircular dome, with trigonometric principles allows you to derive complex relationships, such as \( r = d \frac{2 \sqrt{2}}{\sqrt{3} - 1} \). Solutions often involve steps like rationalizing denominators to reach the simplest form.
These concepts underlie many practical applications, highlighting the beautiful intersection of geometry and algebra via trigonometry.