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Chapter 28: Problem 69

If the angles of elevation of the top of a tower from three collinear points \(A, B\) and \(C\), on a line leading to the foot of the tower, are \(30^{\circ}, 45^{\circ}\) and \(60^{\circ}\) respectively, then the ratio, \(A B: B C\), is: (A) \(\sqrt{3}: \sqrt{2}\) (B) \(1: \sqrt{3}\) (C) \(2: 3\) (D) \(\sqrt{3}: 1\)

Short Answer

Expert verified
The ratio is \(\sqrt{3}: 1\) (option D).

Step by step solution

01

Understand the Scenario

Consider a scenario where three points \(A\), \(B\), and \(C\) are collinear and are on a line leading to the base of a tower. The angles of elevation of the top of the tower from these points are \(30^{\circ}\), \(45^{\circ}\), and \(60^{\circ}\) respectively. We need to find the ratio \(AB:BC\).
02

Define the Problem Mathematically

Let the height of the tower be \(h\), and the distance from the base of the tower to point \(B\) be \(x\). The distance from \(A\) to the base is greater than \(x\), and from \(C\) to the base is less than \(x\). The angles of elevation give us the relationships based on trigonometric ratios.
03

Use Trigonometric Ratios

From point \(B\), which has a \(45^{\circ}\) elevation angle, we can say \(\tan 45^{\circ} = \frac{h}{x} = 1\) which implies \(h = x\).
04

Derive Equations for Other Points

For point \(A\), \(\tan 30^{\circ} = \frac{h}{x + AB}\). Since \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\), we have \(\frac{h}{x+AB} = \frac{1}{\sqrt{3}}\). This gives us \(AB = \sqrt{3}x - x = (\sqrt{3} - 1)x\).
05

Derive Equation for Point C

For point \(C\), \(\tan 60^{\circ} = \frac{h}{x - BC}\). Since \(\tan 60^{\circ} = \sqrt{3}\), we have \(\frac{h}{x-BC} = \sqrt{3}\). Solving for \(BC\), \(BC = x - \frac{x}{\sqrt{3}} = x(1 - \frac{1}{\sqrt{3}})\).
06

Calculate the Desired Ratio

Now we find \(AB:BC\). We have \(AB = (\sqrt{3} - 1)x\) and \(BC = x(1 - \frac{1}{\sqrt{3}})\). Simplifying \(BC = x \cdot \frac{\sqrt{3} - 1}{\sqrt{3}}\). The ratio \(AB:BC = \frac{\sqrt{3} - 1}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \sqrt{3}:1\).
07

Confirm the Ratio Conclusion

Verify the arithmetic and logic used in previous steps, ensuring no calculation errors. The desired ratio is \(\sqrt{3}: 1\).

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Key Concepts

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

Angles of Elevation
The angle of elevation is a crucial concept in trigonometry that refers to the angle between the horizontal line from an observer and the line of sight to an object above that horizontal line. Imagine standing several feet away from a towering skyscraper. When you look up at its peak, you're forming an angle of elevation.
  • An observer's eye level marks the horizontal line.
  • The line from their eyes to the top of the tower shows the angle of elevation.
Mathematically, this angle helps us relate the height of an object and the distance from its base using trigonometric functions.
Applications: Using angles of elevation, you can effectively determine heights of buildings, mountains, and even monitor the slope of ramps. This concept plays a fundamental role in navigation and architectural design.
Trigonometric Ratios
Trigonometric ratios are essential tools in understanding the relationships between angles and sides of right-angled triangles. When we talk about trigonometric ratios, we usually refer to sine (\( ext{sin}\)), cosine (\( ext{cos}\)), and tangent (\( ext{tan}\)). These ratios are invaluable in solving various mathematical problems.
  • Sine: Opposite side over hypotenuse, \( ext{sin} \theta = \frac{ ext{Opposite}}{ ext{Hypotenuse}} \).
  • Cosine: Adjacent side over hypotenuse, \( ext{cos} \theta = \frac{ ext{Adjacent}}{ ext{Hypotenuse}} \).
  • Tangent: Opposite side over adjacent side, \( ext{tan} \theta = \frac{ ext{Opposite}}{ ext{Adjacent}} \).
In the context of angles of elevation, the tangent is particularly useful. It helps relate the height of an object and its horizontal distance from an observer.

For example, in our problem:
  • For point B with a 45-degree angle, \( an 45^{\circ} = \frac{h}{x} = 1\) implies that the height and horizontal distance are equal.
These relationships enable us to create and solve equations effectively, defining distances and heights with efficacy.
Collinear Points
The concept of collinear points arises when three or more points lie on the same straight line. In any geometric problem, understanding if points are collinear helps simplify the analysis. Imagine aligning several dots on a sheet of paper so they form a straight pathway鈥攖hese are collinear.

In this trigonometric context, having points A, B, and C be collinear is vital for solving the problem of angles of elevation.
  • As all three points lie on a single straight path to the base of the tower, the path can be mathematically analyzed seamlessly.
  • This alignment helps us create the right equations to find out desired ratios or distances.
The condition of points being collinear means we can apply straightforward trigonometric principles, such as assuming a single line without any curves or deviations, which would otherwise complicate calculations.

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Most popular questions from this chapter

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 \(A\) sq. metre, then the area of the hexagon is (A) \(\frac{3 \sqrt{3}}{2} A\) sq. metres (B) \(\frac{3 \sqrt{3}}{\pi} A\) sq. metres (C) \(\frac{3 \sqrt{3}}{2} \frac{A}{\pi}\) sq. metres (D) none of these

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 person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is \(60^{\circ}\) and when he retires 40 meter away from the tree the angle of elevation becomes \(30^{\circ} .\) The breadth of the river is (A) \(20 \mathrm{~m}\) (B) \(30 \mathrm{~m}\) (C) \(40 \mathrm{~m}\) (D) \(60 \mathrm{~m}\)

An isosceles triangle of wood of base \(2 a\) and height \(h\) is placed with its base on the ground and vertex directly above. The triangle faces the sun whose altitude is \(30^{\circ}\). Then the tangent of the angle at the apex of the shadow is (A) \(\frac{2 h a}{\sqrt{3}}\) (B) \(\frac{2 h a \sqrt{3}}{3 h^{2}-a^{2}}\) (C) \(\frac{a^{2}+h^{2}}{2 \sqrt{3}}\) (D) \(\frac{2 a h \sqrt{3}}{3 h^{2}+a^{2}}\)

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
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