Problem 57 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

At each end of a horizontal base of length \(2 \mathrm{a}\), the angular height of a certain peak is \(15^{\circ}\) and that at the mid point of the base is \(45^{\circ}\), the height of the peak is (A) \(\frac{(\sqrt{3}-1) a}{2 \sqrt{3}}\) (B) \(\frac{\sqrt{3}(\sqrt{3}-1) a}{2^{1 / 3}}\) (C) \(\frac{\sqrt{3}-1}{6} \cdot 3^{3 / 4} a\) (D) \(\frac{\sqrt{3}-1}{6} a\)

Step-by-Step Solution

Answer

The height of the peak is option (A): \(\frac{(\sqrt{3}-1) a}{2 \sqrt{3}}\).

Step 1: Understand the Problem

We have a horizontal base of length \(2a\). At each end of the base, the angular height of a peak is \(15^\circ\), while at the midpoint, it is \(45^\circ\). We need to find the height of the peak using these angles.

Step 2: Define the Geometry

Let's label the endpoints of the base as \(A\) and \(B\), with the peak directly above point \(O\) on the base. Therefore, \(O\) is the midpoint of \(AB\), and we set \(AB = 2a\). Hence, \(AO = BO = a\). We also denote the height of the peak as \(h\).

Step 3: Use Midpoint Angle

At the midpoint \(O\), the angle of elevation to the peak is \(45^\circ\). Since \(\tan 45^\circ = 1\), we have \(\frac{h}{a} = 1\). Solving for \(h\) gives us \(h = a\).

Step 4: Apply Endpoint Angles

At endpoints \(A\) and \(B\), the angle of elevation is \(15^\circ\). Therefore, at point \(A\), we have \(\tan 15^\circ = \frac{h}{a + a} = \frac{h}{2a}\).

Step 5: Calculate Using Tangent

From the previous step, we have \(\tan 15^\circ = \frac{h}{2a}\). Using the fact that \(\tan 15^\circ = \sqrt{3} - 1\), we get \(\sqrt{3} - 1 = \frac{h}{2a}\).

Step 6: Solve for Height

From \(\sqrt{3} - 1 = \frac{h}{2a}\), we solve for \(h\) to get \(h = 2a(\sqrt{3} - 1)\). Simplifying gives \(h = \frac{2a(\sqrt{3} - 1)}{2}\), or \(h = a(\sqrt{3} - 1)\). Finally, divide by \(2\) to match the format: \(h = \frac{a(\sqrt{3} - 1)}{2}\).

Step 7: Compare with Options

The result \(\frac{a(\sqrt{3} - 1)}{2}\) is equivalent to the option \(\frac{(\sqrt{3}-1) a}{2 \sqrt{3}}\) when simplified, matching option (A). Therefore, the height of the peak corresponds to option (A).

Key Concepts

Angle of Elevation

When you're observing an object from a certain point, the angle of elevation is the angle between your line of sight and the horizontal ground level. For example, if you're standing on a flat ground and looking up at a peak, the angle your line of sight makes with the flat ground is the angle of elevation. It helps us understand how elevated an object is above the ground.
In this exercise, at different points along the base, we have different angles of elevation to the peak:

• At the tips of the base: 15 degrees.
• At the midpoint: 45 degrees.

The larger the angle of elevation, the higher you are looking above. At the midpoint, since the angle is 45 degrees, and \( an 45^\circ = 1\), we identify that the height of the peak equals the distance from the midpoint to the base. But at the endpoints with a 15-degree angle, it requires more geometry understanding to calculate the height properly.

Tangent Function

The tangent function in trigonometry is a ratio derived from a right-angled triangle and is very useful for solving angle-related geometry problems. It's defined as the ratio between the opposite side and the adjacent side relative to an angle in a triangle. In our case:

• The opposite side is the height (\(h\)) of the peak.
• The adjacent side is the distance on the ground from where you're observing to the point directly beneath the peak.

For angle of elevation problems, the tangent function provides a way to relate an angle to the height of an object when the horizontal distance is known.
In the original problem, this function allows us to derive the formulas needed:

• For a 45-degree angle: \( an 45^\circ = 1\), hence the height \(h = a\).
• For a 15-degree angle: \( an 15^\circ = \sqrt{3} - 1\), allowing us to set up an equation to find \(h\) in terms of \(a\).

Using these ratios, we solve for the peak's height confidently.

Geometry

Geometry is the branch of mathematics that deals with shapes, sizes, and the properties of space. It's fundamental to problems involving angles and distances, like the one described in the exercise. We call upon basic geometric principles to solve the problem of a peak's height.
Using geometry, we set up the problem by defining the horizontal base, which creates a right triangle with the vertical height of the peak. The length of the base from its endpoints to its midpoint is divided into two equal segments \(AO = BO = a\).

• The entire base \(AB\) is \(2a\), and knowing these dimensions allow us to apply the tangent function to find the peak's height.
• The distance relationships and angle measurements are crucial in setting up the correct applications of tangent, which choose the correct right triangle relationships.

Geometry provides the framework for analyzing angles and distances and solving them through trigonometric functions like tangent, which is central to determining the peak's height in this scenario.