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Chapter 28: Problem 25

From a point on a hill-side of constant inclination, the angle of elevation of the top of a flagstaff on its summit is observed to be \(\alpha\) and \(a\) metre nearer the top of the hill, it is \(\beta\). If \(h\) is the height of the flagstaff, the inclination of the hill to the horizontal is (A) \(\sin ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (B) \(\cos ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (C) \(\tan ^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin (\beta-\alpha)}\right)\) (D) none of these

Short Answer

Expert verified
The inclination of the hill is (A) \(\sin^{-1}\left(\frac{a \sin \alpha \sin \beta}{h \sin(\beta-\alpha)}\right)\).

Step by step solution

01

Set Up and Understand the Scenario

Visualize the problem by imagining a triangle formed by the observer's point, the top of the hill, and the top of the flagstaff. As seen by the observer, this triangle has angles at the base related to angles of elevation \(\alpha\) and \(\beta\) when closer by distance \(a\).
02

Introduce Relevant Trigonometric Identities

Recall the trigonometric identity for the angle of elevation \( \tan(\theta) = \frac{\text{Op. Side}}{\text{Adj. Side}} \). Knowing \(\tan\) for both scenarios, express each in terms of the height \(h\) of the flagstaff and the distances involved.
03

Equation Formulation

For the initial observation point, we say \( \tan(\alpha) = \frac{h}{d} \), where \(d\) is the distance along the hill's surface to the flagstaff base. For a point \(a\) meters closer, the tan equation changes to \( \tan(\beta) = \frac{h}{d-a} \).
04

Relate the Two Equations

Subtract the two equations to eliminate \(h\), leading to \( \tan(\beta) (d - a) = \tan(\alpha) d \). Simplifying, it shows geometric progression involving \( \tan \).
05

Solve for Inclination

Use trigonometric identities sin and cos to manipulate into desired form, focusing on sin of the angle only, such that \(d = \frac{a \sin \alpha \sin \beta}{\sin(\beta - \alpha)} \), revealing how the surface inclination yields from derived trigonometric simplifications.
06

Evaluate the Choices

The original trigonometric manipulation confirms that choice \(A\) is the appropriate expression for the hill's inclination calculated as the arcsin.

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Key Concepts

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

Angle of Elevation
The angle of elevation is a basic trigonometric concept used to describe the angle between the horizontal plane and the line of sight from an observer to an object above them. It is often used in problems involving heights and distances.
When standing on the side of a hill and looking up at a flagstaff, the angle of elevation refers to how steeply you must tilt your eyes upward to see the top of the flagstaff.
  • This angle helps determine the relationship between the distance to the object and its height.
  • It is usually measured using a device like a theodolite or calculated using trigonometric functions like tangent.
  • The angle of elevation plays a crucial role in navigational problems and architectural design.
Understanding the angle of elevation is key to solving problems involving inclined planes, as it allows us to create right triangles to apply trigonometric identities appropriately.
Inclination of Hill
The inclination of the hill represents the angle made between the horizontal ground and the surface of the hill itself. It is an important concept in understanding how steep or gradual the hill is.
The inclination directly affects how we perceive objects that are situated on the hill, particularly when it comes to calculating angles of elevation.
  • This concept is often dealt with in terms of its sine, cosine, or tangent values, as these trigonometric functions help us quantify the inclination angle.
  • Knowing the inclination allows engineers and architects to design structures that account for gravity's force on the inclined surface.
  • The steeper the hill, the higher the inclination angle, affecting how objects are viewed or approached when climbing.
Understanding the hill's inclination enables us to apply specific trigonometric identities, such as \( an(\alpha) = \frac{h}{d} \), to solve practical problems.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domain. These identities are the backbone of solving trigonometric problems, such as those involving angles of elevation and inclines.
In the given exercise, trigonometric identities are used to relate different angles and distances on an inclined plane.
  • The tangent identity, \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\), helps to solve problems related to triangles and inclination.
  • Sine and cosine identities allow the manipulation of equations to determine unknown quantities like slope or hill inclination.
  • The ability to subtract and equate these identities helps to isolate and find specific angles, such as the arc sine in the solution.
Mastering these identities simplifies solving many geometric and trigonometric problems, making them essential tools in mathematics and various fields.

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Most popular questions from this chapter

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

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 a cloud from a point \(h\) metres above the surface of a lake is \(\theta\) and the angles of depression of its reflection is \(\phi\). Then the height of the cloud is (A) \(\frac{h \sin (\phi+\theta)}{\sin (\phi-\theta)}\) (B) \(\frac{h}{\sin (\phi-\theta)}\) (C) \(h \tan (\phi-\theta)\) (D) \(\frac{h \sin (\phi-\theta)}{\sin (\phi+\theta)}\)

A person standing on the ground observes the angle of elevation of the top of a tower to be \(30^{\circ}\). On walking a distance a in a certain direction he finds the elevation of the top to be the same as before. He then walks a distance \(5 \mathrm{a} / 3\) at right angles to his former direction, and finds that the elevation of the top has doubled. The height of the tower is (A) \(\sqrt{5 / 6} a\) (B) \(\sqrt{85 / 48} a\) (C) \(\sqrt{6 / 5} a\) (D) \(\sqrt{48 / 85} a\)

The angular elevation of tower \(C D\) at a point \(A\) due south of it is \(60^{\circ}\) and at a point \(B\) due west of \(\mathrm{A}\), the elevation is \(30^{\circ}\). If \(A B=3 \mathrm{~km}\), the height of the tower is (A) \(2 \sqrt{3} \mathrm{~km}\) (B) \(2 \sqrt{6} \mathrm{~km}\) (C) \(\frac{3 \sqrt{3} \mathrm{~km}}{2}\) (D) \(\frac{3 \sqrt{6} \mathrm{~km}}{4}\)
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