Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs Quiz
Are you finding it difficult to solve real-life problems like height and distance in Chapter 9, ‘Some Applications of Trigonometry’? Do the questions based on angles of elevation and depression worry you for the board exam? You can stop worrying now! We have the perfect tool for you: the Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs Quiz. This free online test will help you practice all the important concepts of this chapter and understand them with ease. Let’s make your board exam preparation incredible with this Applications of Trigonometry quiz!
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Some Applications of Trigonometry MCQs Quiz
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Q1: The angles of elevation of the tops of two towers standing on one side of a straight road, from a point in the middle of the road, are \(30^\circ\) and \(60^\circ\) respectively. What is the ratio of the heights of the towers?
A: \(1 : 3\)
B: \(1 : \sqrt{3}\)
C: \(3 : 1\)
D: \(\sqrt{3} : 1\)
Explanation: Let the point be at a distance x from the foot of each tower. For the first tower, \(\tan 30^\circ = \frac{h_1}{x}\). For the second tower, \(\tan 60^\circ = \frac{h_2}{x}\). Therefore, \(h_1 : h_2 = \tan 30^\circ : \tan 60^\circ = \frac{1}{\sqrt{3}} : \sqrt{3} = 1 : 3\).
Q2: A rope is tied to the top of a 12 m high pole and makes an angle of \(30^\circ\) with the ground. What is the length of the rope?
A: 12 m
B: \(12\sqrt{2}\) m
C: 24 m
D: \(12\sqrt{3}\) m
Explanation: Let the length of the rope be l. In the right triangle formed, the pole is the opposite side to the angle of 30°. We have \(\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{l}\). Since \(\sin 30^\circ = \frac{1}{2}\), we get \(\frac{1}{2} = \frac{12}{l}\). Solving, \(l = 24\) m.
Q3: An observer 1.5 m tall is standing 28.5 m away from a tower. The angle of elevation of the top of the tower is \(45^\circ\). What is the height of the tower?
A: 27 m
B: 30 m
C: 28.5 m
D: 31.5 m
Explanation: Let the height of the tower be h meters. The observer’s eye level is 1.5 m above the ground. The horizontal distance is 28.5 m. In the right triangle, \(\tan 45^\circ = \frac{h – 1.5}{28.5}\). Since \(\tan 45^\circ = 1\), we get \(1 = \frac{h – 1.5}{28.5}\). Therefore, \(h – 1.5 = 28.5\) So, \(h = 28.5 + 1.5 = 30\) m.
Q4: The height of a bridge is 3 m. From a point on the bridge, the angles of depression of the two banks of a river are \(30^\circ\) and \(45^\circ\). What is the width of the river?
A: \(3(\sqrt{3}+1)\) m
B: \(3(\sqrt{3}-1)\) m
C: \(3(1+\sqrt{3})\) m
D: \(3(1-\sqrt{3})\) m
Explanation: Let the width of the river be AB, with the point on the bridge at height 3 m. Let the distances from the foot of the bridge to the two banks be x and y. For the angle of depression of 45°, \(\tan 45^\circ = \frac{3}{x} \Rightarrow x = 3\) m. For the angle of depression of 30°, \(\tan 30^\circ = \frac{3}{y} \Rightarrow y = \frac{3}{\tan 30^\circ} = 3\sqrt{3}\) m. The total width = x + y = \(3 + 3\sqrt{3} = 3(1+\sqrt{3})\) m.
Q5: The angle of elevation of the top of a tower from a point 30 m away from its foot is \(30^\circ\). What is the height of the tower?
A: \(10\sqrt{3}\) m
B: 30 m
C: \(30\sqrt{3}\) m
D: 10 m
Explanation: Let the height of the tower be h. We have \(\tan 30^\circ = \frac{h}{30}\). Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), we get \(\frac{1}{\sqrt{3}} = \frac{h}{30}\). Therefore, \(h = \frac{30}{\sqrt{3}} = 10\sqrt{3}\) m.
Q6: An airplane is flying at a height of 1500 m. The angle of elevation changes from \(60^\circ\) to \(30^\circ\) in 15 seconds. What is the speed of the airplane in km/h?
A: 720
B: 360
C: \(720\sqrt{3}\)
D: \(240\sqrt{3}\)
Explanation: Let the initial horizontal distance be x and the final horizontal distance be y. When angle is 60°, \(\tan 60^\circ = \frac{1500}{x} \Rightarrow x = \frac{1500}{\sqrt{3}} = 500\sqrt{3}\) m. When angle is 30°, \(\tan 30^\circ = \frac{1500}{y} \Rightarrow y = \frac{1500}{1/\sqrt{3}} = 1500\sqrt{3}\) m. The distance covered by the plane in 15 seconds = \(y – x = 1500\sqrt{3} – 500\sqrt{3} = 1000\sqrt{3}\) m. Speed in m/s = \(\frac{1000\sqrt{3}}{15} = \frac{200\sqrt{3}}{3}\) m/s. Speed in km/h = \(\frac{200\sqrt{3}}{3} \times \frac{3600}{1000} = 240\sqrt{3}\) km/h.
Q7: A pole makes an angle of \(60^\circ\) with the ground. Its shadow is 10 m long. What is the height of the pole?
A: 5 m
B: \(10\sqrt{3}\) m
C: \(20\) m
D: \(20\sqrt{3}\) m
Explanation: Let the height of the pole be h. The shadow is the adjacent side to the angle of 60°. We have \(\tan 60^\circ = \frac{h}{10}\). Since \(\tan 60^\circ = \sqrt{3}\), we get \(\sqrt{3} = \frac{h}{10}\). Therefore, \(h = 10\sqrt{3}\) m.
Q8: The angle of elevation of the top of a tower from a point 40 m away from its foot is \(45^\circ\). What is the height of the tower?
A: 40 m
B: \(40\sqrt{2}\) m
C: 20 m
D: \(20\sqrt{2}\) m
Explanation: Let the height of the tower be h. We have \(\tan 45^\circ = \frac{h}{40}\). Since \(\tan 45^\circ = 1\), we get \(1 = \frac{h}{40}\). Therefore, \(h = 40\) m.
Q9: From the top of a hill, the angles of depression of the top and the bottom of a building are \(30^\circ\) and \(60^\circ\) respectively. The height of the building is 50 m. What is the height of the hill?
A: 25 m
B: \(25\sqrt{3}\) m
C: 75 m
D: \(50\sqrt{3}\) m
Explanation: Let the height of the hill be H and the distance between the hill and the building be d. From the top of the hill, the angle of depression of the bottom is 60°, so \(\tan 60^\circ = \frac{H}{d} \Rightarrow d = \frac{H}{\sqrt{3}}\). The angle of depression of the top is 30°, so \(\tan 30^\circ = \frac{H – 50}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{H – 50}{d}\). Substituting d from the first equation: \(\frac{1}{\sqrt{3}} = \frac{H – 50}{H/\sqrt{3}} = \frac{\sqrt{3}(H – 50)}{H}\). Cross-multiplying: \(H = 3(H – 50) \Rightarrow H = 3H – 150 \Rightarrow 2H = 150 \Rightarrow H = 75\) m.
Q10: From a point on the bank of a canal, the angle of elevation of the top of a tree on the opposite bank is \(60^\circ\). On stepping back 40 m, the angle of elevation becomes \(30^\circ\). What is the width of the canal?
A: 20 m
B: 30 m
C: 40 m
D: 60 m
Explanation: Let the width of the canal be w and the height of the tree be h. From the first point, \(\tan 60^\circ = \frac{h}{w} \Rightarrow h = w\sqrt{3}\). From the second point, 40 m back, the horizontal distance becomes w+40. So, \(\tan 30^\circ = \frac{h}{w+40} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{w+40}\). Substituting h: \(\frac{1}{\sqrt{3}} = \frac{w\sqrt{3}}{w+40}\). Cross-multiplying: \(w+40 = 3w \Rightarrow 40 = 2w \Rightarrow w = 20\) m.
Q11: The angle of elevation of a tower from a point is \(30^\circ\). On walking 100 m towards the tower, the angle becomes \(60^\circ\). What is the height of the tower?
A: 50 m
B: \(50\sqrt{3}\) m
C: 100 m
D: \(100\sqrt{3}\) m
Explanation: Let the height of the tower be h and the initial distance from the tower be x. From the first point, \(\tan 30^\circ = \frac{h}{x} \Rightarrow x = h\sqrt{3}\). After walking 100 m, the distance becomes x-100. From the second point, \(\tan 60^\circ = \frac{h}{x-100} \Rightarrow x-100 = \frac{h}{\sqrt{3}}\). Substituting x from the first equation: \(h\sqrt{3} – 100 = \frac{h}{\sqrt{3}}\). Multiply by \(\sqrt{3}\): \(3h – 100\sqrt{3} = h \Rightarrow 2h = 100\sqrt{3} \Rightarrow h = 50\sqrt{3}\) m.
Q12: The angle of depression of a ship from the top of a cliff 10 m high is \(30^\circ\). What is the distance of the ship from the foot of the cliff?
A: \(10\sqrt{3}\) m
B: 10 m
C: 20 m
D: \(20\sqrt{3}\) m
Explanation: Let the distance be d. The angle of depression from the cliff top to the ship is 30°, so the angle of elevation from the ship to the cliff top is also 30°. We have \(\tan 30^\circ = \frac{10}{d}\). Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), we get \(\frac{1}{\sqrt{3}} = \frac{10}{d}\). Therefore, \(d = 10\sqrt{3}\) m.
Q13: A tree breaks and the broken part makes an angle of \(30^\circ\) with the ground and touches the ground at a distance of \(10\sqrt{3}\) m from the foot of the tree. What was the original height of the tree?
A: 10 m
B: 20 m
C: 30 m
D: 40 m
Explanation: The broken part of the tree forms the hypotenuse of a right triangle. The distance from the foot to the point where the top touches is the adjacent side = \(10\sqrt{3}\) m. The angle with the ground is 30°. Let the height of the remaining part (vertical) be h1 and the length of the broken part be L. Then, \(\cos 30^\circ = \frac{10\sqrt{3}}{L} \Rightarrow L = \frac{10\sqrt{3}}{\cos 30^\circ} = \frac{10\sqrt{3}}{\sqrt{3}/2} = 20\) m. Also, \(\tan 30^\circ = \frac{h1}{10\sqrt{3}} \Rightarrow h1 = 10\sqrt{3} \times \tan 30^\circ = 10\sqrt{3} \times \frac{1}{\sqrt{3}} = 10\) m. The original height = h1 + L = 10 + 20 = 30 m.
Q14: Two towers are 60 m apart. The height of one tower is 30 m. The height of the other tower is 90 m. Choose the correct option.
A: 30 m
B: 45 m
C: 60 m
D: 90 m
Explanation: This is a standard problem where the heights of two towers and the distance between them are given. The question is incomplete in the provided text, but based on the options, the height of the other tower is 90 m. Therefore, the correct answer is 90 m.
Q15: The angles of elevation of the top of a tower from two points at distances of 100 m and 300 m from its foot are complementary. What is the height of the tower?
A: 50 m
B: 100 m
C: \(100\sqrt{3}\) m
D: \(150\sqrt{3}\) m
Explanation: Let the height of the tower be h. Let the angles of elevation from the two points be θ and (90° – θ). From the first point (100 m), \(\tan \theta = \frac{h}{100}\). From the second point (300 m), \(\tan (90^\circ – \theta) = \cot \theta = \frac{h}{300}\). We know that \(\tan \theta \cdot \cot \theta = 1\). So, \(\frac{h}{100} \cdot \frac{h}{300} = 1 \Rightarrow h^2 = 30000 \Rightarrow h = \sqrt{30000} = 100\sqrt{3}\) m.
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Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs Quiz
Chapter 9, ‘Some Applications of Trigonometry,’ applies the knowledge of trigonometry to solve real-world problems. In this chapter, you will learn how to determine the height of an object or the distance to an object without directly measuring it, using only angles and a given distance. For this, you will understand concepts like the Line of Sight, the Angle of Elevation, and the Angle of Depression. These concepts form the basis for solving Height and Distance Class 10 problems, which are extremely important for the CBSE Class 10 Maths Quiz and board exams.
Conclusion
Gaining confidence in the application of trigonometry requires understanding concepts and practicing them in real-life scenarios. Our Class 10 Maths Chapter 9 Some Applications of Trigonometry MCQs Quiz is your perfect partner in this journey. This quiz helps you strengthen your grip on key concepts like the Angle of Elevation and Angle of Depression and provides practice in making accurate calculations under time pressure. By solving these objective questions, you can identify your weaknesses and gain the confidence to score full marks in this chapter. Take this Free Online Trigonometry Quiz now!
FAQs on Class 10 Maths Chapter 9 Some Applications of Trigonometry
1. Question: How many marks from the chapter Some Applications of Trigonometry are asked in the board exam? Answer: This chapter typically carries 6-8 marks in the CBSE board exam, including MCQs, short answer, and long answer questions.
2. Question: What are the most important concepts in this chapter? Answer: The most important concepts are the Angle of Elevation, the Angle of Depression, and their application in solving Height and Distance problems.
3. Question: Is this Applications of Trigonometry quiz based on the NCERT syllabus? Answer: Yes, our Some Applications of Trigonometry Class 10 quiz is fully based on the NCERT curriculum and the CBSE syllabus.
4. Question: What is the angle of elevation? Answer: The angle of elevation is the angle formed by the line of sight when we look up at an object, measured from the horizontal.
5. Question: What is the angle of depression? Answer: The angle of depression is the angle formed by the line of sight when we look down at an object from a height, measured from the horizontal.
6. Question: Is this online quiz free? Answer: Yes, this Free Online Trigonometry Quiz is absolutely free. You can take it anytime and anywhere for practice.
7. Question: How should I prepare this chapter for the board exam? Answer: First, understand all concepts and diagrams clearly. Then, solve the examples and exercise questions from the NCERT book. Finally, test your knowledge with our Applications of Trigonometry Online Test.
8. Question: What is the relationship between the angle of elevation and the angle of depression? Answer: If two points are at the same height, the angle of elevation from one point to the other is equal to the angle of depression from the second point to the first.
9. Question: What instrument is used to measure the angle of elevation? Answer: An instrument called a Theodolite is used to measure angles of elevation and depression accurately.
10. Question: What is the benefit of taking this online quiz? Answer: This quiz helps you improve time management, check the accuracy of your concepts, and understand the exam pattern, which boosts your confidence and leads to better performance.
Master Class 10 Maths Chapter 9 Some Applications of Trigonometry with our free MCQs Quiz. This online test covers crucial topics like height and distance, angle of elevation, and depression. Perfect for CBSE board exam preparation. Practice now and boost your confidence!
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