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Chapter 28: Problem 15

\(A B C D\) 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) \(48 \sqrt{3} s q . m\)

Short Answer

Expert verified
(A) 48 鈭�2 sq. m.

Step by step solution

01

Understand the Problem

We have a rectangular field ABCD with a lamp post of height 12 m at corner A. We need to find the area of the field using the angles of elevation to the lamp's top from corners B and C which are 60掳 and 45掳 respectively.
02

Set Up Coordinate System

Place the field on a coordinate plane with point A at the origin (0,0). Let AB be on the x-axis with B at (x,0) and AD on the y-axis with D at (0,y). Thus C is at (x,y).
03

Apply Trigonometry at Point B

The angle of elevation from B to the top of the lamp post is 60掳. Therefore, the height of the lamp post (12m) equals the opposite side of the right triangle with the run from A to B as the adjacent side:\[\tan(60^{\circ}) = \frac{12}{x}\]Solving, we get:\[x = \frac{12}{\tan(60^{\circ})} = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{3} = 4\sqrt{3}\]
04

Apply Trigonometry at Point C

For the angle of elevation from C, which is 45掳, the height is equal to the run along either axis (since tan45掳 = 1):\[\tan(45^{\circ}) = \frac{12}{z}\]Where z is the distance from A to C (hypotenuse along AD):\[\Rightarrow z = 12\]
05

Calculate Length AD

Now, using A, C coordinates as (x,y) where x = 4鈭�3 from previous results and z = hypotenuse:\[z^2 = x^2 + y^2\Rightarrow 12^2 = (4\sqrt{3})^2 + y^2\]Continuing:\[144 = 48 + y^2\Rightarrow y^2 = 96\Rightarrow y = 4\sqrt{6}\]
06

Calculate Area of the Field

ABCD is a rectangle with sides AB and AD. Using x = 4鈭�3 and y = 4鈭�6 we find the area:\[\text{Area} = x \cdot y = (4\sqrt{3}) \cdot (4\sqrt{6})\]Solve the product:\[= 16 \sqrt{18} \Rightarrow = 48\sqrt{2} \text{ square meters}\].
07

Match with Options

Finally, compare the derived area 48鈭�2 sq. meters with the given options. It matches with option (A) 48鈭�2 sq. meters.

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Key Concepts

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

Rectangular Field Geometry
Understanding the layout of a rectangular field is essential in many mathematical problems. A rectangle is a four-sided polygon with opposite sides being equal in length. This makes it a quadrilateral with angles that total 360 degrees, having four right angles, each measuring 90 degrees. Simply put, opposite sides will always run parallel to each other unless stated otherwise in problems involving fields.
  • Opposite sides equal: In the rectangle ABCD, the length AB equals CD, and AD equals BC.
  • Right angles present: Each corner angle measures 90 degrees, making the internal angles consistent.
In problems involving geometry of rectangular fields, these properties help set constraints, ensuring logical coherence in problem-solving. This helps in realizing that if you know one side of a rectangle, you can often find its disproportionate or congruent counterpart if needed.
Angles of Elevation
Angles of elevation are crucial in practical geometry and trigonometry calculations. They are the angle formed by the line of sight from an observer located at a lower point to an object situated at a higher point, relative to the horizontal line.
  • Understanding 60掳 elevation: When the angle of elevation is 60掳 from point B to the top of the lamp post, it creates a height-to-base ratio using trigonometric identities.
  • Exploring 45掳 elevation: At 45掳, angles provide intuitive balance in geometric formations since the opposite side (height) and adjacent side (base) are equal, reflecting symmetry.
These angles are often utilized to calculate heights and distances using the tangent function, which is beneficial when trying to relate various positional elements geometrically, such as in this exercise.
Area Calculation
To determine the area of a rectangular field, comprehend the underlying formula: the product of its length and width. In simpler terms, it is the multiplication of two adjacent sides.
  • The area formula: Area = Length 脳 Width
  • Specific context: Using the given lengths, for example when Length AB = 4鈭�3 and Width AD = 4鈭�6, the application becomes effortless.
Calculation involves multiplication being extended through roots or variables based on the initial evidences and solving for depicting logical results after simplification. Therefore, this setup transforms into an arithmetic problem revolving around foundational mathematics principles.
Right Triangle Properties
Right triangle properties are instrumental in solving height and distance problems, especially with angles of elevation. Key traits include one angle fixed at 90 degrees and the remaining two supplementary angles.
  • Right Angles: They facilitate trigonometric ratios, like sine, cosine, and tangent, fundamental for computing other triangle components.
  • Hypotenuse Length: Represents the side opposite the right angle, longest in the right triangle category, linking adjacent and opposite sides.
  • Tan Function Usage: Tangent ratio is extremely practical in elevation angle problems, as seen from pitchers like tan(60掳) and tan(45掳) in problem calculations.
With these dynamics, practitioners can evaluate and analyze spatial configurations and dimensions under geometrical constraints ensuring successful mathematical problem resolutions.

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Most popular questions from this chapter

The angle of elevation of a tower from a point \(\mathrm{A}\) due south of it is \(\mathrm{x}\) and from a point \(\mathrm{B}\) due east of \(\mathrm{A}\) is \(\mathrm{y}\). If \(\mathrm{AB}=1\), then the height \(\mathrm{h}\) of the tower is given by (A) \(\frac{l}{\sqrt{\cot ^{2} y-\cot ^{2} x}}\) (B) \(\frac{l}{\sqrt{\tan ^{2} y-\tan ^{2} x}}\) (C) \(\frac{2 l}{\sqrt{\cot ^{2} y-\cot ^{2} x}}\) (D) none of these

Assertion: A pole of length \(h\) stands inside a triangular plot \(A B C\) and subtends equal angles \(\alpha\) at its vertices, then \(2 h \cos \alpha \sin A=a \sin \alpha\). Reason: For circumscribed radius \(R\) of a \(\triangle A B C\), \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R .\)

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 \(\mathrm{h}\) 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

A stationary balloon is observed from three points \(\mathrm{A}\), \(\mathrm{B}\) and \(\mathrm{C}\) on the plane ground and is found that its angle of elevation from each point is \(\alpha\). If \(\angle \mathrm{ABC}=\beta\) and \(\mathrm{AC}=\mathrm{b}\), the height of the balloon is (A) \(\frac{b}{2} \tan \alpha \operatorname{cosec} \beta\) (B) \(\frac{b}{2} \tan \alpha \sin \beta\) (C) \(\frac{b}{2} \cot \alpha \operatorname{cosec} \beta\) (D) \(\frac{b}{2} \cot \alpha \sin \beta\)

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) \(\frac{2 a h \sqrt{3}}{3 h^{2}+a^{2}}\)
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