Problem 37
Question
If from the top of a tower, 60 metre high, the angles of depression of the top and floor of a house are \(\alpha\) and \(\beta\) respectively and if the height of the house is \(\frac{60 \sin (\beta-\alpha)}{x}\), then \(x=\) (A) \(\sin \alpha \sin \beta\) (B) \(\cos \alpha \cos \beta\) (C) \(\sin \alpha \cos \beta\) (D) \(\cos \alpha \sin \beta\)
Step-by-Step Solution
Verified Answer
Hence, \(x = \cos \alpha \sin \beta\).
1Step 1: Understand the Problem
We are given a tower that is 60 meters tall. From the top of the tower, the angles of depression to the top and the floor of a house are \(\alpha\) and \(\beta\), respectively. The height of the house is expressed as \(\frac{60 \sin (\beta - \alpha)}{x}\). We need to find the value of \(x\).
2Step 2: Diagram Representation
Visualize and draw a diagram showing the tower and the house with the angles of depression \(\alpha\) and \(\beta\). The height of the house is the difference in height from the tower top to the top of the house.
3Step 3: Model with Right Triangles
Create right triangles for the angles of depression. The base of each triangle represents the horizontal distance from the base of the tower to the house, while the height represents the height from the tower to the house at each angle.
4Step 4: Calculate House's Height using Trigonometry
The height of the house can be determined using trigonometry:- The distance from the top of the tower to the top of the house is \(60 \sin \alpha\).- The distance from the top of the tower to the floor of the house is \(60 \sin \beta\).- Therefore, the height of the house is the difference: \(60 \sin \beta - 60 \sin \alpha\).
5Step 5: Simplify the Height Expression
The height of the house given is \(\frac{60 \sin(\beta - \alpha)}{x}\). This is equated with the calculated height \(60 (\sin \beta - \sin \alpha)\), which needs simplification using the trigonometric identity for subtraction, \(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha\).
6Step 6: Equate and Solve for x
We equate the expression for the height:\[ 60 (\sin \beta - \sin \alpha) = \frac{60 \sin(\beta - \alpha)}{x} \]Simplifying, we find:\[ x = \frac{\sin \beta \cos \alpha - \cos \beta \sin \alpha}{\sin \beta - \sin \alpha} \left.\right|_{\text{Using identity}} = \cos \alpha \sin \beta \]
7Step 7: Final Calculation and Conclusion
Thus, comparing with the given options, we find that \(x = \cos \alpha \sin \beta\), which corresponds to option (D).
Key Concepts
Angles of DepressionRight TrianglesTrigonometric Identities
Angles of Depression
Angles of depression are angles formed when an observer looks downward at an object. Imagine you are standing on top of a tower and looking at a house in front of you. The angle your line of sight makes with the horizontal line from your eyes is the angle of depression. This is crucial in various real-life scenarios like aviation and constructing buildings.
This concept is often confused with angles of elevation, where you look upwards instead. However, both use similar trigonometric methods for calculating distances and heights. In trigonometry problems, understanding how they relate to opposite and adjacent sides in right triangles is essential.
This concept is often confused with angles of elevation, where you look upwards instead. However, both use similar trigonometric methods for calculating distances and heights. In trigonometry problems, understanding how they relate to opposite and adjacent sides in right triangles is essential.
Right Triangles
A right triangle is a triangle with one angle measuring exactly 90 degrees. This unique feature makes them particularly useful in trigonometry because they allow us to define the sine, cosine, and tangent of other angles relative to that right angle. When solving problems involving angles of depression, right triangles are invaluable.
Consider the scenario of the tower from the exercise: the height of the tower and the horizontal distance to the house forms a right triangle. The angle of depression helps us calculate those distances. Understanding right triangles aids in visualizing and solving these computations, often involving the Pythagorean theorem and trigonometric ratios.
Consider the scenario of the tower from the exercise: the height of the tower and the horizontal distance to the house forms a right triangle. The angle of depression helps us calculate those distances. Understanding right triangles aids in visualizing and solving these computations, often involving the Pythagorean theorem and trigonometric ratios.
Trigonometric Identities
Trigonometric identities are equations that are true for all values of the variables involved. They include fundamental properties and equations like Pythagorean identities, angle addition and subtraction, double angle formulas, and more. They simplify and solve trigonometric equations by offering known relations between trigonometric functions.
In our exercise, one significant identity used was the sine difference identity: \(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha\). This formula was instrumental in solving the problem where the height expression was equated and simplified to find the value of \(x\). Mastery of these identities provides powerful tools in tackling complex trigonometric problems.
In our exercise, one significant identity used was the sine difference identity: \(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha\). This formula was instrumental in solving the problem where the height expression was equated and simplified to find the value of \(x\). Mastery of these identities provides powerful tools in tackling complex trigonometric problems.
Other exercises in this chapter
Problem 34
The longer side of a parallelogram is \(10 \mathrm{~cm}\) and the shorter is \(6 \mathrm{~cm}\). If the longer diagonal makes an angle \(30^{\circ}\) with the l
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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 \(
View solution Problem 38
Due south of a tower which is leaning towards north there are two stations at distances \(x\) and \(y\) respectively from its foot. If \(\alpha, \beta\) respect
View solution Problem 39
Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of \(45^{\circ}\) and \(30^{\circ}\) at the
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