Problem 2 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

A vertical tower subtends an angle of \(60^{\circ}\) at a point on the same level as the foot of the tower. On moving 100 \(\mathrm{m}\) further from the first point in line with the tower, it subtends an angle of \(30^{\circ}\) at the point. Find the height of the tower.

Step-by-Step Solution

Answer

The height of the tower is \(50 \sqrt{3}\) meters.

Step 1: Identify Given Information

In the given problem, the tower is the side opposite to the angles \(60^{\circ}\) and \(30^{\circ}\) respectively. If we can find the length of the adjacent side, we can calculate the height of the tower using the tangent of the angles which is the ratio of the side opposite to the angle to the adjacent side.

Step 2: Establish First Equation

In the triangle formed by the tower and the first reference point, we have the angle \(60^{\circ}\). Let the height of the tower be \(h\), and the distance of this point from the tower be \(d\). Using the definition of tangent, we have \[tan(60^{\circ}) = \frac{h}{d}\]. Or, rearranging to express \(d\) in relation to \(h\), we have \[d = \frac{h}{tan(60^{\circ})}\]. We know \(tan(60^{\circ})\) is \(\sqrt{3}\), so our equation becomes \[d = \frac{h}{\sqrt{3}}\]

Step 3: Establish Second Equation

Similarly, in the triangle formed by the tower and the second reference point, we have the angle \(30^{\circ}\). The distance of this point from the tower will be \(d+100\) (as we moved 100 meters further) and the height \(h\). Using the definition of tangent, we have \[tan(30^{\circ}) = \frac{h}{d+100}\]. \[ d+100 =\frac{h}{tan(30^{\circ})}\] We know that \(tan(30^{\circ})\) is \(\frac{1}{\sqrt{3}}\), so substituting gives us \[d+100 =h \sqrt{3}\].

Step 4: Solve Equations Simultaneously

By equating the expressions for \(d\) from the first and the second equation, we get \[\frac{h}{\sqrt{3}}+100 = h \sqrt{3}\] Dividing the entire equation by \(\sqrt{3}\) will give us \(h + 100 \sqrt{3}= 3h\). This simplifies to \(2h = 100 \sqrt{3}\), hence the height \(h = \frac{100 \sqrt{3}}{2} = 50 \sqrt{3} meters\).\]

Key Concepts

Angle of Elevation

In trigonometry, the angle of elevation is a fundamental concept that describes the angle between the horizontal line of sight and an object above the observer's eye level. For instance, if you are looking upwards at a tower, the angle your line of sight makes with the horizontal is the angle of elevation.

• This angle is crucial in tasks that involve calculating heights and distances when direct measurement is not feasible.
• Real-world applications include determining the height of buildings, mountains, and even finding the altitude of an aircraft.

For example, if a tower subtends an angle of elevation of \(60^{\circ}\) from a point on level ground, this implies the observer is looking upwards at this angle in reference to a horizontal line.
Understanding and calculating this angle helps in utilizing trigonometric functions like tangent effectively for solving practical problems like the one in our problem.

Tangent Function

The tangent function is one of the primary trigonometric functions and is defined in a right-angled triangle as the ratio of the side opposite the angle to the side adjacent to the angle. Mathematically, for an angle \(\theta\), it is represented as:

• \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \)
• This function helps solve height and distance problems especially when you know an angle and one side of the triangle.

For our problem, using the tangent function with the angles \(60^{\circ}\) and \(30^{\circ}\), we can derive equations to find the height of the tower. We used the relationships \( \tan(60^{\circ}) = \sqrt{3}\) and \( \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \) to set up the equations.

Height and Distance Problems

Height and distance problems in trigonometry involve determining unknown lengths like height or distance using known angles and trigonometric functions. These problems often utilize the angle of elevation for calculations.

• Such questions usually form one or more right-angled triangles where either the height (vertical side) or the distance (horizontal side) has to be determined.
• The tangent function is commonly used, especially when dealing with problems involving two points forming two different angles from the object's top or bottom.

Our exercise exemplifies a typical height and distance problem. It required setting up simultaneous equations using tangent values from two distinct points, with each forming different angles of elevation. Solving these equations yielded the required height of the tower, demonstrating practical use of trigonometric principles.

Trigonometric Ratios

Trigonometric ratios are the relationships between the sides of a right triangle and its angles. These include sine, cosine, and tangent, which are essential for solving various problems involving triangles. Here's a brief overview:

• Sine is the ratio of the length of the opposite side to the hypotenuse: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\)
• Cosine is the ratio of the adjacent side to the hypotenuse: \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• Tangent is the ratio of opposite over adjacent sides: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)

These ratios are invaluable in calculating distances, angles, and other measurements in both theoretical and real-world scenarios. In the given problem, understanding the tangent ratio allowed us to calculate the unknown height of the tower using the given angles of \(60^{\circ}\) and \(30^{\circ}\), bringing out the practical application of trigonometric ratios.