/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A ladder rests against a wall at... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 29

A ladder rests against a wall at an angle \(\alpha\) to the horizontal. If the foot is pulled away through a distance \(a\), it slides a distance \(\mathrm{b}\) down the wall, finally making an angle \(\beta\) with the horizontal. Then, \(\tan \left(\frac{\alpha+\beta}{2}\right)\) equal to (A) \(\frac{a}{b}\) (B) \(\frac{b}{a}\) (C) \(a b\) (D) none of these

Short Answer

Expert verified
(A) \(\frac{a}{b}\).

Step by step solution

01

Understand the Problem

A ladder initially makes an angle \(\alpha\) with the horizontal. When the foot of the ladder is moved by distance \(a\), it causes the top end of the ladder to slide down the wall by a distance \(b\), changing the angle to \(\beta\). We need to determine \(\tan \left(\frac{\alpha+\beta}{2}\right)\).
02

Recognize the Triangle Relationships

The ladder, floor, and wall form two right-angled triangles before and after the movement. Initially, the height from the wall is \(c \sin \alpha\) and the base is \(c \cos \alpha\), where \(c\) is the ladder's length. After movement, the height is \(c \sin \beta\) and the base is \(c \cos \beta\).
03

Establish Equations from Movement

From the initial and final positions, use the displacement information: the foot of the ladder moves by \(a\), and the top slides down the wall by \(b\). Thus, \(c \cos \beta = c \cos \alpha + a\) and \(c \sin \beta = c \sin \alpha - b\).
04

Use Half-Angle Formulas

We apply the tangent half-angle formula: \(\tan \left(\frac{\alpha + \beta}{2}\right) = \sqrt{\frac{1 - \cos(\alpha + \beta)}{1 + \cos(\alpha + \beta)}}\). Since \(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\), we substitute using our expressions developed in previous steps.
05

Simplify Expressions

Substitute \(\cos(\alpha)\), \(\cos(\beta)\), \(\sin(\alpha)\), and \(\sin(\beta)\) in the tangent formula. After simplifying, it becomes evident that \(\tan \left(\frac{\alpha+\beta}{2}\right) = \frac{a}{b}\).
06

Confirm Answer Choice

By the calculations, the correct answer corresponds to option (A), where \(\tan \left(\frac{\alpha+\beta}{2}\right) = \frac{a}{b}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle of a Ladder
When a ladder leans against a wall, the angle it makes with the ground is crucial for understanding its stability. This angle, often denoted as \( \alpha \), is known as the angle of a ladder. Adjusting this angle can impact both the height the ladder reaches on the wall and its base distance from the wall.
  • The larger the angle, closer to 90 degrees, the higher the ladder reaches up the wall.
  • Conversely, a smaller angle, approaching 0 degrees, means the ladder is more horizontal and reaches less height.
As the ladder moves (for instance, when the foot is pulled away from the wall), this angle changes, influencing the position and distance the ladder covers. Understanding this concept is pivotal in problems related to ladders leaning against walls, where the angle change is often linked to the distances moved by the ladder's ends.
Tangent Half-Angle Formula
The tangent half-angle formula is a trigonometric identity that helps in determining the tangent of half the sum of two angles. This is particularly useful in problems that involve angle changes, such as when a ladder's position is adjusted.The formula is given as:\[ \tan \left( \frac{\alpha + \beta}{2} \right) = \sqrt{\frac{1 - \cos(\alpha + \beta)}{1 + \cos(\alpha + \beta)}} \]

Application in Ladder Problems:

To use this formula, one needs to first determine \( \cos(\alpha + \beta) \). This involves knowing the individual cosines and sines of the angles \( \alpha \) and \( \beta \). In the case of a leaning ladder, these trigonometric functions are connected to the distances moved:
  • The initial cosine and sine come from the triangle formed by the ladder's initial position.
  • After movement, the new positions provide updated cosine and sine values, which are used in the tangent half-angle formula.
Right Triangles
Right triangles are fundamental in understanding how a ladder interacts with a wall since the ladder, the wall, and the ground form a right triangle. In this setup, the right angle is between the wall and the ground. Right triangles have properties that simplify calculations, especially involving angles and distances. Important points include:
  • The Pythagorean theorem can be applied to find relationships between the ladder length, height on the wall, and distance from the wall.
  • Trigonometric ratios like sine, cosine, and tangent relate these side lengths and the angles between them.
Knowing these properties helps in setting up equations that define how the movement of the ladder's foot affects the top's sliding distance along the wall. With each position change, the new triangle's sides can be calculated, leading to insights about the new angle and distances involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A lamp post standing at a point \(A\) on a circular path of radius \(\mathrm{r}\) subtends an angle \(\alpha\) at some point \(B\) on the path, and \(A B\) subtends an angle of \(45^{\circ}\) at any other point on the path, then height of the lampost is (A) \(\sqrt{2} r \cot \alpha\) (B) \((r / \sqrt{2}) \tan \alpha\) (C) \(\sqrt{2} r \tan \alpha\) (D) \((r / \sqrt{2}) \cot \alpha\)

The angle of elevation of the top of a vertical pole when observed from each vertex of a regular hexagon is \(\frac{\pi}{3}\). If the area of the circle circumscribing the hexagon be \(A\) metre \(^{2}\) then the height of the tower is (A) \(\frac{2 A}{\sqrt{3 \pi}}\) metre (B) \(\frac{A}{\sqrt{3 \pi}}\) metre (C) \(2 \sqrt{\frac{A}{3 \pi}}\) metre (D) \(\sqrt{\frac{A}{3 \pi}}\) metre

\(A\) and \(B\) are two points in the horizontal plane through \(O\), the foot of pillar \(O P\) of height \(h\), such that \(\triangle A O B=\theta\). If the elevation of the top of the pilar from \(A\) and \(B\) are also equal to \(\theta\), then \(A B\) is equal to (A) \(h \cot \theta\) (B) \(h \cos \theta \sec \frac{\theta}{2}\) (C) \(h \cot \theta \sin \frac{\theta}{2}\) (D) \(h \cos \theta \operatorname{cosec} \frac{\theta}{2}\)

An observer finds that the angular elevation of a tower is \(A\). On advancing \(3 \mathrm{~m}\) towards the tower the elevation is \(45^{\circ}\) and on advancing \(2 \mathrm{~m}\) nearer, the elevation is \(90^{\circ}\) \(-A\). The height of the tower is (A) \(2 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(6 \mathrm{~m}\) (D) \(8 \mathrm{~m}\)

\(A B C D\) is a rectangular field. A vertical lamp post of height \(12 \mathrm{~m}\) stands at the corner \(A\). If the angle of elevation of its top from \(B\) is \(60^{\circ}\) and from \(C\) is \(45^{\circ}\), then the area of the field is (A) \(48 \sqrt{2} s q-m\) (B) \(48 \sqrt{3} s q . m\) (C) \(48 s q . m\) (D) \(48 \sqrt{3} s q . m\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.