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\(\mathrm{ABC}\) is a triangular park with \(\mathrm{AB}=\mathrm{AC}=100\) metres. \(\mathrm{A}\) vertical tower is situated at the mid-point of \(B C\). If the angles of elevation of the top of the tower at \(\mathrm{A}\) and \(\mathrm{B}\) are \(\cot ^{-1}(3 \sqrt{2})\) and \(\operatorname{cosec}^{-1}(2 \sqrt{2})\) respectively, then the height of the tower (in metres) is: [April 10, 2019 (I)] (a) \(\frac{100}{3 \sqrt{3}}\) (b) \(10 \sqrt{5}\) (c) 20 (d) 25

Short Answer

Expert verified
The height of the tower is \( \frac{100}{3\sqrt{3}} \) meters.

Step by step solution

01

Understanding the Geometry

We are given a triangle ABC in which AB = AC = 100 meters and a tower at the midpoint of BC. Let D be the midpoint of BC. We need to calculate the height of the tower AD.
02

Calculate Angle at A

The given angle of elevation at A is \( \cot^{-1}(3 \sqrt{2}) \). Therefore, the cotangent of angle θ at A is \( 3 \sqrt{2} \). In terms of tangent function, \( \tan(\theta) = \frac{1}{3 \sqrt{2}} \).
03

Calculate Angle at B

The given angle of elevation at B is \( \operatorname{cosec}^{-1}(2 \sqrt{2}) \). Therefore, the cosecant of angle at B is \( 2 \sqrt{2} \). In terms of sine function, \( \sin(\phi) = \frac{1}{2 \sqrt{2}} \).
04

Express Heights in Terms of Trigonometric Ratios

From the trigonometric identities, for angle at A, height \( h = AD = \tan(\theta) \times AD \). For angle at B, height \( h = AD = \sin(\phi) \times BD \).
05

Equate and Solve for Height

At A, \( \tan(\theta) = \frac{1}{3 \sqrt{2}} \); \( h = \frac{1}{3 \sqrt{2}} \times 100 \). At B, \( \sin(\phi) = \frac{1}{2 \sqrt{2}} \); \( h = \frac{1}{2 \sqrt{2}} \times 50 \). Solving both expressions provides the height of the tower.
06

Final Calculation

Calculate the expression for h using the relations derived. \( h = \frac{100}{6\sqrt{3}} \) or simplify further to get \( h = \frac{100}{3\sqrt{3}} \).

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Key Concepts

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

Triangle Geometry
Triangle Geometry deals with the properties and relations of triangles. The given exercise involves a triangle ABC, where the sides AB and AC are both 100 meters long, forming an isosceles triangle. The special aspect of this geometry is that a vertical tower is located at the midpoint of side BC.

In this setup, because the tower stands at the midpoint of BC, it creates two equal halves of the base, allowing us to explore symmetrical properties. With AB and AC equal, we often find that segments intersecting through the midpoint create right triangles. This allows us to simplify calculations using basic trigonometric ratios, knowing that triangles formed share equal characteristics, due to symmetry.

Additionally, understanding the distinction between different types of triangles, such as right, isosceles, or equilateral, helps in identifying which trigonometric ratios or laws are best suited to find missing lengths or angles.
Trigonometric Identities
Trigonometric identities are essential tools for solving problems involving angles and sides of triangles. In this exercise, we use different trigonometric identities to express angles of elevation seen in the triangle.
  • The **cotangent** function involved at angle A is presented as \( \cot^{-1}(3 \sqrt{2}) \). It can be converted to tangent because \( \tan(\theta) = \frac{1}{\text{cot}(\theta)} \), allowing us to work with ratios of adjacent to opposite sides more effectively. Hence, \( \tan(\theta) = \frac{1}{3 \sqrt{2}} \).
  • Meanwhile, the **cosecant** function at angle B follows the identity \( \operatorname{cosec}^{-1}(2 \sqrt{2}) \). It relates to the sine function by \( \sin(\phi) = \frac{1}{\text{cosec}(\phi)} \), giving \( \sin(\phi) = \frac{1}{2 \sqrt{2}} \).
These transformations allow us to express the height of the tower using simple algebraic manipulations based on the known ratios of triangle sides. Knowing how to pivot between these identities can greatly simplify solving complex triangle problems.
Angle of Elevation
The Angle of Elevation is a critical concept in trigonometry and practical applications like surveying and navigation. It is the angle between the horizontal plane and the line of sight upward to an object above the horizontal plane.In this exercise, two angles of elevation are provided: one at point A and another at point B. These angles are key to determining the height of the tower. - At point A, the angle of elevation translates through trigonometric identities to a tangent value, helping calculate the distance upwards from the ground to the tower's peak by using the formula: \( h = \tan(\theta) \times \text{adjacent side length} \).

- Similarly, at point B, the angle being expressed through the sine function relates directly to how high the tower climbs relative to this point's horizontal level, allowing the calculation of height using: \( h = \sin(\phi) \times \text{adjacent side length} \).Understanding how these angles influence side lengths and height verticality is pivotal for accurate distance and height calculations. The concept of angle elevation thus bridges geometric dimensions with practical, real-world measurements.

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Most popular questions from this chapter

For a regular polygon, let \(r\) and \(R\) be the radii of the inscribed and the circumscribed circles. A false statement among the following is (a) There is a regular polygon with \(\frac{r}{R}=\frac{1}{\sqrt{2}}\) (b) There is a regular polygon with \(\frac{r}{R}=\frac{2}{3}\) (c) There is a regular polygon with \(\frac{r}{R}=\frac{\sqrt{3}}{2}\) (d) There is a regular polygon with \(\frac{r}{R}=\frac{1}{2}\)

In a triangle, the sum of lengths of two sides is \(x\) and the product of the lengths of the same two sides is \(\mathrm{y}\). if \(x^{2}-c^{2}=y\), where \(c\) is the length of the third side of the triangle, then the circumradius of the triangle is : [Jan. 11,2019 (I)] (a) \(\frac{3}{2} y\) (b) \(\frac{c}{\sqrt{3}}\) (c) \(\frac{c}{3}\) (d) \(\frac{y}{\sqrt{3}}\)

Let \(\mathrm{A}(3,0,-1), \mathrm{B}(2,10,6)\) and \(\mathrm{C}(1,2,1)\) be the vertices of a triangle and \(\mathrm{M}\) be the midpoint of \(\mathrm{AC}\). If G divides \(\mathrm{BM}\) in the ratio, \(2: 1\), then \(\cos (\angle \mathrm{GOA})\) (O being the origin) is equal to: [April 10, 2019 (I)] (a) \(\frac{1}{2 \sqrt{15}}\) (b) \(\frac{1}{\sqrt{15}}\) (c) \(\frac{1}{6 \sqrt{10}}\) (d) \(\frac{1}{\sqrt{30}}\)

The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is [2003] (a) \(\frac{a}{4} \cot \left(\frac{\pi}{2 n}\right)\) (b) \(a \cot \left(\frac{\pi}{n}\right)\) (c) \(\frac{a}{2} \cot \left(\frac{\pi}{2 n}\right)\) (d) \(a \cot \left(\frac{\pi}{2 n}\right)\).

The angle of elevation of the top of a vertical tower from a point \(\mathrm{P}\) on the horizontal ground was observed to be \(\alpha .\) After moving a distance 2 metres from \(\mathrm{P}\) towards the foot of the tower, the angle of elevation changes to \(\beta\). Then the height (in metres) of the tower is: [Online April 11, 2014] (a) \(\frac{2 \sin \alpha \sin \beta}{\sin (\beta-\alpha)}\) (b) \(\frac{\sin \alpha \sin \beta}{\cos (\beta-\alpha)}\) (c) \(\frac{2 \sin (\beta-\alpha)}{\sin \alpha \sin \beta}\) (d) \(\frac{\cos (\beta-\alpha)}{\sin \alpha \sin \beta}\)

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