Problem 3 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Find the height of a tower when it is found that on walking \(80 \mathrm{~m}\) towards it along a horizontal line through its base, the angular elevation of its top changes from \(30^{\circ}\) and \(60^{\circ}\).

Step-by-Step Solution

Answer

The height of the tower is \(40\sqrt{3} + 40\) meters.

Step 1: Develop the diagram and the first triangle

Visualize the scenario. There's a person observing the top of a tower and then moving towards it. Two right-angled triangles can be formed here. In the first triangle (let's label it ABC), with the initial position of the observer as A, the base of tower as B and the top of tower as C. Here, angle C is 90 degrees, angle BAC (angle of elevation) is 30 degrees. Assume h to be the height of the tower (BC).

Step 2: Apply trigonometry to the first triangle

Use the definition of the tangent of an angle in a right triangle: \(\tan(\theta) = \frac {opposite}{adjacent}\). Apply this to angle BAC in triangle ABC: \(\tan(30^{\circ}) = \frac {h}{AB}\). Solve for AB: \(AB = h / \tan(30^{\circ}) = h \sqrt{3}\).

Step 3: Develop the second triangle and apply trigonometry

When the observer moves 80 meters closer to the tower forming triangle A'B'C. Here, angle A'BC (angle of elevation) is 60 degrees and A'B (distance of observer from tower after moving) is AB-80. Now apply trigonometry to this triangle. We have: \(\tan(60^{\circ}) = \frac {h}{A'B} = \frac {h}{h \sqrt{3} - 80}\)

Step 4: Solve the equation

Solving the equation \(\sqrt{3} = \frac {h \sqrt{3}}{h \sqrt{3} - 80}\) will provide the value of h. Multiply both sides by h \sqrt{3} - 80 to clear the denominator and get: \( h \sqrt{3} - 80 = h\) . Simplify this equation to get: \( h\sqrt{3} - h = 80\). Factor out 'h': \( h (\sqrt{3} - 1) = 80\). Solving for h, we get: \[ h = \frac {80}{\sqrt{3} - 1} \]

Step 5: Rationalize the denominator

It's typically preferred to not have a sqrt in the denominator, so we multiply numerator and denominator by conjugate: \( \sqrt{3} + 1 \). Thus the height of the tower is: \[ h = \frac {80 (\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac {80 (\sqrt{3} + 1)}{3 -1} = 40 \sqrt{3} + 40 ~m \]

Key Concepts

Angle of Elevation

The angle of elevation is an important concept in trigonometry and geometry.
It is the angle formed by the horizontal line from the observer's eye to an object above the observer.
For example, when you look at the top of a tower, the angle between your line of sight and the horizontal ground is the angle of elevation. Typically, this angle is measured from the ground, or directly from the line parallel to the ground.

• The angle of elevation is crucial in problems involving heights and distances.
• As one gets closer to the object (like a tower), the angle of elevation increases.
• It helps in calculating unknown distances or heights when other values, like distances, are known.

Understanding the angle of elevation assists in breaking down complex trigonometric problems into simpler right-triangle problems. By visualizing these angles, one can better grasp how distances and heights relate to each other.

Tangent Function

In trigonometry, the tangent function is associated with angles in right-angled triangles. The tangent of an angle in such a triangle is defined as the ratio of the length of the opposite side to the length of the adjacent side.
In mathematical terms: \[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]

• The tangent function is especially useful in problems involving heights, like determining the height of a building or tower given a specific angle.
• It is the preferred function when the opposite and adjacent sides of a right triangle are the primary values in question.
• The tangent of common angles like \(30^{\circ}\), \(45^{\circ}\), and \(60^{\circ}\) often shows up in exercises, making it easier with standard ratios like \(\tan(30^{\circ}) = \frac{1}{\sqrt{3}}\).

By applying the tangent function, you can derive essential measurements that might not be directly apparent, especially in right-angled triangles involving unknown heights or lengths.

Right-Angled Triangles

Right-angled triangles are a cornerstone of trigonometry.
One of the angles in a right-angled triangle is always exactly \(90^{\circ}\). This property leads to the uniquely valuable relationships between the triangle's sides and angles as captured by trigonometric functions, such as sine, cosine, and tangent.

• In a right-angled triangle, the side opposite the right angle is called the hypotenuse.
• The other two sides are referred to as the "opposite" and "adjacent" sides in relation to the angle of interest.
• Right-angled triangles allow for direct calculation of angles and side lengths using trigonometry.

Understanding the basic composition and properties of right-angled triangles is vital, as it simplifies solving numerous real-world problems. Once you grasp how these triangles work, applying trigonometric functions becomes more intuitive.

Rationalizing the Denominator

Rationalizing the denominator is a process used in mathematics to eliminate square roots (or other irrational numbers) from the denominator of a fraction.
It's a preferred form because it makes the expression clearer and simpler to work with.

• To rationalize, multiply the numerator and the denominator by the conjugate of the denominator.
• For example, if we have a denominator such as \(\sqrt{3} - 1\), its conjugate is \(\sqrt{3} + 1\).
• Multiplying these conjugates results in a rational number because \((\sqrt{3} - 1)(\sqrt{3} + 1)\) simplifies to \(3 - 1\), which equals 2.

By rationalizing the denominator, expressions become easier to understand and use, particularly in further calculations, ensuring the results remain accurate and practical.