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Chapter 28: Problem 66

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that \(\mathrm{AB}\) ( \(=\) a) subtends an angle of \(60^{\circ}\) at the foot of the tower, and the angle of elevation of the top of the tower from \(\mathrm{A}\) or \(\mathrm{B}\) is \(30^{\circ} .\) The height of the tower is [2007] (A) \(\frac{2 a}{\sqrt{3}}\) (B) \(2 a \sqrt{3}\) (C) \(\frac{a}{\sqrt{3}}\) (D) \(a \sqrt{3}\)

Short Answer

Expert verified
The height of the tower is \(a\).

Step by step solution

01

Understand the Problem

Imagine a circular park with a tower at the center. Points \(A\) and \(B\) are on the edge such that line \(AB\) subtends a \(60^{\circ}\) angle at the tower's base, and the angle of elevation from both \(A\) and \(B\) to the top of the tower is \(30^{\circ}\). We need to find the height of the tower.
02

Apply Geometry Concepts

Use trigonometric concepts. Given \( \angle AOB = 60^{\circ} \), the triangle \( OAB \) forms an equilateral triangle with side \( a \) because the subtended angle and known angle of elevation suggest symmetry.
03

Use Trigonometry for Tower Height

From \(A \) or \(B\), the height \( h \) can be found using \( \tan(30^{\circ}) \). This implies \( \tan(30^{\circ}) = \frac{h}{\frac{a}{\sqrt{3}}} \) given a \(30^{\circ}\) angle from points \(A\) or \(B\) to the top, and a distance from center \( \frac{a}{\sqrt{3}} \).
04

Solve the Equation

The tangent of \(30^{\circ}\) is \( \frac{1}{\sqrt{3}} \). Set up the equation: \( \frac{1}{\sqrt{3}} = \frac{h}{\frac{a}{\sqrt{3}}} \). Solving for \(h\) gives:\[ h = \frac{a}{\sqrt{3}} \cdot \sqrt{3} = a \]
05

Verify the Answer

Re-examining the calculations and logic, the solution is valid with the given formula and values.

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Key Concepts

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

Geometric Concepts
Geometric concepts form the foundation for understanding spatial relationships. They involve properties and attributes of figures and shapes. In this exercise, we deal with a circular park and a tower positioned at its center. Consider an equilateral triangle, a shape where all sides are of equal length and angles are equal.
  • When you have a circle with a center point, any triangle formed by connecting points on the circle鈥檚 boundary (like in our exercise) involves significant symmetry.
  • In such a scenario, the line from the center to any point on the circle is a radius. This concept is crucial when angles and distances around the circumference are involved.
Understanding geometric symmetry such as that seen here helps in calculating unknown dimensions efficiently using angles and known distances.
Angle of Elevation
The angle of elevation is a geometric concept used to describe the angle between the horizontal plane and the line of sight. When you gaze upward to a point on a higher object (like the top of a tower), the angle between your sightline and the ground is the angle of elevation.
  • In this exercise, both points \(A\) and \(B\) have an angle of elevation of \(30^\circ\) to the top of the tower.
  • This uniform angle suggests a repeated trigonometric relationship and symmetry, which simplifies the calculation of the tower鈥檚 height.
  • Tangents are used frequently to solve these problems since \(\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}\).
Using these principles, especially for common angles like 30, 45, or 60 degrees, can simplify calculations significantly.
Equilateral Triangle
An equilateral triangle is a type of triangle where each side length is equal, and each interior angle measures \(60^\circ\). In our exercise, the triangle \(OAB\) highlights this.
  • Since the line \(AB\) subtends an angle of \(60^\circ\) at the foot of the tower, it forms such an equilateral triangle with the tower鈥檚 base at the center.
  • This geometric property makes each side of the triangle equally \(a\), and they form relative constraints simplifying trigonometric solutions.
  • Due to this, symmetry ensures certain distances or heights will relate uniformly, which allows for straightforward algebraic resolution, like calculating the tower's height from the given data.
Understanding these triangles provides insight into more complex geometry and aids solving various mathematical problems.

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Most popular questions from this chapter

\(\mathrm{AB}\) is a vertical pole with \(\mathrm{B}\) at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point \(\mathrm{C}\) on the ground is \(60^{\circ} .\) He moves away from the pole along the line \(\mathrm{BC}\) to a point \(\mathrm{D}\) such that \(\mathrm{CD}=7 \mathrm{~m}\). From D the angle of elevation of the point \(\mathrm{A}\) is \(45^{\circ}\). Then the height of the pole is (A) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} m\) (B) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m\) (C) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) m\) (D) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}\)

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) \((r / \sqrt{2}) \cot \alpha\)

A tower \(A B\) leans towards west making an angle \(\alpha\) with the vertical. The angular elevation of \(B\), the top most point of the tower is \(\beta\), as observed from a point \(C\) due east of \(A\) at a distnace \(d\) from \(A\). If the angular elevation of \(B\) from a point due east of \(C\) at a distance \(2 d\) from \(C\) is \(\gamma\), then (A) \(2 \tan \alpha=2 \cot \beta-\cot \gamma\) (B) \(2 \tan \alpha=3 \cot \beta-\cot \gamma\) (C) \(\tan \alpha=\cot \beta-\cot \gamma\) (D) none of these

\(A B\) is a vertical pole with \(B\) at the ground level and \(A\) at the top. A man finds that the angle of elevation of the point \(A\) from a certain point \(C\) on the ground is \(60^{\circ} .\) He moves away from the pole along the line \(B C\) to a point \(D\) such that \(C D=7 \mathrm{~m} .\) From \(D\) the angle of elevation of the point \(A\) is \(45^{\circ}\). Then the height of the pole is (A) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} \mathrm{~m}\) (B) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) \mathrm{m}\) (C) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) \mathrm{m}\) (D) \(\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}\)

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) \(\frac{h \sin (\phi-\theta)}{\sin (\phi+\theta)}\)
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