Problem 9 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The angle of elevation of the top of vertical tower from a point \(A\) on the horizontal ground is found tobe \(\frac{\pi}{4}\). From ' \(A\) ' a man walks \(10 \mathrm{~m}\) up a path sloping at an angle \(\pi / 6\). After this the slope becomes steeper and after walking up another \(10 \mathrm{~m}\), the man reaches the top of the tower. Find the distance of ' \(A\) ' from the foot of the tower.

Step-by-Step Solution

Answer

The distance of point ' A ' from the foot of the tower is \(20\sqrt{3} m\).

Step 1: Identify given values

The angle of elevation from A to the top of the tower is \(\frac{\pi}{4}\). The man first walks 10 m up a path with a slope angle of \(\frac{\pi}{6}\), then walks another 10 meters up a steeper slope to the top of the tower.

Step 2: Apply trigonometric functions on the first slope

In the right-angled triangle formed by the man's path, the horizontal ground, and the line from the starting point to the end of the 10 m walk, we can use the tangent of the slope angle. Let's denote the horizontal distance from A to the foot of the tower by \(x\), and the height gained on the first walk by \(h1\). Then \(tan(\pi/6)=h1/x\), so \(h1=x \cdot tan(\pi/6)\). The total height after the first slope is \(10+h1\).

Step 3: Apply trigonometric functions on the total height

The angle of elevation from A to the top of the tower measures the total height of the tower, which is the height gained on the first slope plus 10 m on the second slope. So, \(tan(\pi/4)=(10+h1)/x\), which simplifies to \(10+h1 = x\).

Step 4: Solve for x (the distance of ' A ' from the foot of the tower)

Replace \(h1\) in the equation \(10+h1=x\) using \(h1=x\cdot tan(\pi/6)\) to get \(10+x \cdot tan(\pi/6)=x\). Solving this equation for \(x\), we find that \(x=20\sqrt{3} m\). This is the distance of point ' A ' from the foot of the tower.

Key Concepts

Angle of Elevation

Understanding the angle of elevation is quite straightforward. It is the angle formed between a horizontal line of sight and the line of sight to an object above the horizontal. In our exercise, this means the angle from point A on the ground to the top of the tower. The given angle here is \(\frac{\pi}{4}\). This tells us how steeply we must look upwards to see the top of the tower from point A.

• Angle of elevation is important for solving problems involving heights and distances.
• Every problem involving an upward glance involves an angle of elevation.

In trigonometry and real-world applications, the angle of elevation helps us to use trigonometric functions to find unknown distances or heights.

Tangent Function

The tangent function is vital in trigonometry and specifically in this tower problem. The tangent of an angle in a right triangle gives the ratio of the opposite side to the adjacent side. For the angle \(\frac{\pi}{6}\), the tangent function helps determine how much the man ascended vertically. This is because: \[ \tan \left( \frac{\pi}{6} \right) = \frac{\text{Height of path}}{\text{Base of path}} \] Applying this to our problem, we find the height gained after walking 10 meters on a path with a slope angle of \(\frac{\pi}{6}\).

• Tangent functions are used to solve for one side when an angle and another side are known.
• They help us relate steepness of slopes or grades to physical sizes in 2D problems.

These functions thus act as a bridge between angles and side lengths.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are crucial tools in mathematics. They allow us to solve various geometry problems involving angles and distances. In this problem, we particularly use the tangent function.
A trigonometric function of an angle translates this angle into a numeric value based on a right triangle's side ratios. This helps solve complex shape sizes in simple equations. For angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) involved in this problem, the tangent function yields required heights.

• Trigonometric functions bring an angle's abstract measures to measurable lengths.
• They allow us to convert geometric configurations into algebraic forms useful for calculations.

Whether it's elongating distances like walking paths or unseen heights like towers, trigonometry makes these known.

Solving Equations

In the exercise, solving equations is the major part. After framing equations using angle of elevation and tangent function, we calculate for unknown parts effectively. This involves algebraic manipulation following clear steps.
To determine the horizontal distance to the tower, set up an equation using known angles and derived heights. Replacing \(h_1 = x \cdot \tan\left(\frac{\pi}{6}\right)\) into \(10 + h_1 = x\), we solve for \(x\): \[ 10 + x \cdot \tan\left(\frac{\pi}{6}\right) = x \]Solving these equations allows finding the distance of point A from the foot of the tower.

• Simplifying and solving equations yields actual measurements from theoretical constructs.
• Understanding math symbol replacements and step-by-step algebraic operations simplifies this task.

Solving equations, hence, ties the whole problem together logically.

IIT JEE Mathematics

The IIT JEE examination challenges students with complex mathematical problems, demanding thorough understanding of trigonometric concepts. This problem is a classic example of how such exams integrate real-world scenarios with mathematical logic. The test favors not just computational skills but also ability to interpret and synthesize different mathematical principles.

• Solving such problems requires a mix of theoretical knowledge and practical application.
• Being well-versed with core trigonometric identities plays a key role.

Studying for IIT JEE mathematics involves:

• Recognizing standard problem types, like angles in elevation or depression.
• Practicing area-specific problems to ensure adept problem-solving techniques.
• Developing proficiency in quick, accurate calculations without error.

Mastering these can help excel in the IIT JEE and similar mathematical challenges, equipping students with critical analytical skills.