Welcome dear students! Today we are going to learn about Some Applications Of Trigonometry from Class 10 Maths. In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying about some ways in which trigonometry is used in the life around you. Let us consider Figure 9.1. In this diagram, a student stands on the ground and looks up at the top of a minar. The line AC drawn from the eye of the student to the top of the minar is called the line of sight. The angle ∠BAC, formed by the line of sight with the horizontal line AB, is called the angle of elevation of the top of the minar from the eye of the student. Thus, the line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of elevation is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, which happens when we raise our head to look at the object, as illustrated in Figure 9.2.
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Now, consider the situation given in Figure 9.3. A girl sitting on a balcony is looking down at a flower pot placed on a stair. In this case, the line of sight is below the horizontal level. The angle so formed by the line of sight with the horizontal is called the angle of depression. Thus, the angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed. Now, you may identify the lines of sight, and the angles so formed in Figure 8.3. Are they angles of elevation or angles of depression?
Let us refer to Figure 9.1 again. If you want to find the height CD of the minar without actually measuring it, you need three pieces of information: the distance DE at which the student is standing from the foot of the minar, the angle of elevation ∠BAC, and the height AE of the student. In the figure, CD equals CB plus BD, where BD equals AE. To find BC, we will use trigonometric ratios of ∠BAC in right triangle ABC. We use tan A equals BC divided by AB, or cot A equals AB divided by BC. Solving gives BC, and adding AE yields the minar height. Now let us explain this process by solving some problems.
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Example 1: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower. Solution: Refer to Figure 9.4. The diagram shows a vertical tower AB and a point C on the ground 15 m from the foot B. ∠ACB is the angle of elevation. We need to determine AB. Triangle ABC is right-angled at B. We choose tan 60°. tan 60° equals AB divided by BC. Substituting values, √3 equals AB divided by 15. Therefore, AB equals 15√3. Hence, the height of the tower is 15√3 m.
Example 2: An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work, as in Figure 9.5. What should be the length of the ladder that she should use, which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? Take √3 equals 1.73. Solution: The diagram shows pole AD of height 5 m. The electrician reaches point B on the pole. BD equals AD minus AB, which is 5 minus 1.3, giving 3.7 m. BC represents the ladder. In right triangle BDC, we use sin 60°. sin 60° equals BD divided by BC. So, √3 divided by 2 equals 3.7 divided by BC. Therefore, BC equals 3.7 multiplied by 2, divided by √3, which is approximately 4.28 m. For the distance DC, we use cot 60°. cot 60° equals DC divided by BD. So, 1 divided by √3 equals DC divided by 3.7. DC equals 3.7 divided by √3, which is approximately 2.14 m. She should place the foot of the ladder 2.14 m from the pole.
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Example 3: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°, as in Figure 9.6. What is the height of the chimney? Solution: In the figure, AB is the chimney, CD is the observer, and ∠ADE is the angle of elevation. Triangle ADE is right-angled at E. DE equals CB, which is 28.5 m. We use tan 45°. tan 45° equals AE divided by DE. So, 1 equals AE divided by 28.5. Therefore, AE equals 28.5 m. The height of the chimney AB equals AE plus BE, which is 28.5 plus 1.5, giving 30 m.
Example 4: From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°, as in Figure 9.7. Find the length of the flagstaff and the distance of the building from point P. Take √3 equals 1.732. Solution: The diagram shows building AB of height 10 m and flagstaff BD. In right triangle PAB, tan 30° equals AB divided by AP. So, 1 divided by √3 equals 10 divided by AP. AP equals 10√3, which is 17.32 m. Let flagstaff DB equal x m. Then AD equals 10 plus x m. In right triangle PAD, tan 45° equals AD divided by AP. So, 1 equals 10 plus x divided by 10√3. Solving gives 10√3 equals 10 plus x, so x equals 10√3 minus 10, which is 10 times (√3 minus 1), equal to 7.32 m. The flagstaff is 7.32 m long.
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Example 5: The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Solution: Refer to Figure 9.8. AB is the tower height h. BC is the shadow length x at 60°. DB is the shadow length x plus 40 at 30°. In right triangle ABC, tan 60° equals AB divided by BC. So, √3 equals h divided by x, meaning h equals √3 times x. In right triangle ABD, tan 30° equals AB divided by BD. So, 1 divided by √3 equals h divided by x plus 40. Substituting h gives 1 divided by √3 equals √3 times x divided by x plus 40. Cross multiplying gives x plus 40 equals 3x. Thus, 2x equals 40, so x equals 20. Substituting back, h equals 20√3 m.
Example 6: The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively, as in Figure 9.9. Find the height of the multi-storeyed building and the distance between the two buildings. Solution: In the diagram, PC is the multi-storeyed building, and AB is the 8 m tall building. PQ is a horizontal line from P. PB is a transversal to parallel lines PQ and BD, so alternate angle ∠PBD equals 30°. Similarly, ∠PAC equals 45°. In right triangle PBD, PD divided by BD equals tan 30° which is 1 divided by √3, so BD equals PD times √3. In right triangle PAC, PC divided by AC equals tan 45° which is 1, so PC equals AC. Since PC equals PD plus DC, and DC equals AB which is 8 m, we have PD plus 8 equals AC. But AC equals BD, so PD plus 8 equals PD times √3. Solving gives PD times (√3 minus 1) equals 8, so PD equals 8 divided by (√3 minus 1). Rationalizing yields PD equals 4 times (√3 plus 1) m. The total height PC equals PD plus DC, which is 4(√3 + 1) plus 8, simplifying to 4√3 plus 12, or 4(√3 + 3) m. Since AC equals PC, the distance between the two buildings is also 4(√3 + 3) m.
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Example 7: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. The bridge is at a height of 3 m from the banks, as in Figure 9.10. Find the width of the river. Solution: In the figure, A and B represent points on the bank on opposite sides, so AB is the river width. P is a point on the bridge at height 3 m, so DP equals 3 m. AB equals AD plus DB. In right triangle APD, ∠A equals 30°. tan 30° equals PD divided by AD. So, 1 divided by √3 equals 3 divided by AD, giving AD equals 3√3 m. In right triangle PBD, ∠B equals 45°. tan 45° equals PD divided by BD which is 1, so BD equals PD, which is 3 m. The width AB equals BD plus AD, which is 3 plus 3√3, or 3 times (1 plus √3) m.
Now we proceed to Exercise 9.1. Question 1: A circus artist is climbing a 20 m long rope, tightly stretched from a vertical pole top to the ground, making a 30° angle with the ground, as in Figure 9.11. Find the pole height. Solution: In right triangle ABC, the rope is hypotenuse AC equals 20 m. The pole is AB. sin 30° equals AB divided by AC. So, 1 divided by 2 equals AB divided by 20. AB equals 10 m.
Question 2: A tree breaks due to a storm and the broken part bends so that the top touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Solution: Let the unbroken part be AB and the broken part be BC. AC equals 8 m. In right triangle ABC, tan 30° equals AB divided by AC. So, 1 divided by √3 equals AB divided by 8, giving AB equals 8 divided by √3. cos 30° equals AC divided by BC. So, √3 divided by 2 equals 8 divided by BC, giving BC equals 16 divided by √3. Total height equals AB plus BC, which is 8 divided by √3 plus 16 divided by √3, giving 24 divided by √3, which simplifies to 8√3 m.
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Question 3: A contractor plans two slides. For younger children, a slide at 1.5 m height inclined at 30°. For elder children, a slide at 3 m height inclined at 60°. Find the lengths. Solution: For the first slide, sin 30° equals 1.5 divided by L1. So, 1 divided by 2 equals 1.5 divided by L1, giving L1 equals 3 m. For the second slide, sin 60° equals 3 divided by L2. So, √3 divided by 2 equals 3 divided by L2, giving L2 equals 6 divided by √3, which simplifies to 2√3 m.
Question 4: The angle of elevation of a tower top from a point 30 m away is 30°. Find the tower height. Solution: tan 30° equals h divided by 30. So, 1 divided by √3 equals h divided by 30. h equals 30 divided by √3, which simplifies to 10√3 m.
Question 5: A kite is flying at 60 m height. The string inclination with the ground is 60°. Find the string length. Solution: sin 60° equals 60 divided by L. So, √3 divided by 2 equals 60 divided by L. L equals 120 divided by √3, which simplifies to 40√3 m.
Question 6: A 1.5 m tall boy stands at some distance from a 30 m tall building. The angle of elevation from his eyes to the top increases from 30° to 60° as he walks towards it. Find the distance walked. Solution: Effective height above his eyes is 30 minus 1.5, which is 28.5 m. Initial distance x1 equals 28.5 divided by tan 30°, which is 28.5√3. Final distance x2 equals 28.5 divided by tan 60°, which is 28.5 divided by √3, simplifying to 9.5√3. Distance walked equals x1 minus x2, which is 28.5√3 minus 9.5√3, giving 19√3 m.
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Question 7: From a point on the ground, angles of elevation to the bottom and top of a transmission tower on a 20 m building are 45° and 60°. Find the tower height. Solution: Let the building height be 20 m and tower height be h. Distance d equals 20 divided by tan 45°, which is 20 m. tan 60° equals 20 plus h divided by 20. So, √3 equals 20 plus h divided by 20. 20√3 equals 20 plus h. h equals 20(√3 minus 1) m.
Question 8: A 1.6 m statue stands on a pedestal. From a point, elevation to pedestal top is 45° and to statue top is 60°. Find pedestal height. Solution: Let pedestal height be h. Distance d equals h divided by tan 45°, which is h. tan 60° equals h plus 1.6 divided by h. So, √3h equals h plus 1.6. h(√3 minus 1) equals 1.6. h equals 1.6 divided by (√3 minus 1), which rationalizes to 0.8(√3 plus 1) m.
Question 9: The angle of elevation of a building top from a tower foot is 30°. The angle of elevation of the tower top from the building foot is 60°. The tower is 50 m high. Find the building height. Solution: Distance d equals 50 divided by tan 60°, which is 50 divided by √3. tan 30° equals h divided by d. So, 1 divided by √3 equals h divided by (50 divided by √3). h equals 50 divided by 3 m.
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Question 10: Two equal poles stand opposite each other on an 80 m road. From a point between them, elevations to the tops are 60° and 30°. Find pole height and distances. Solution: Let height be h. Distances from the point are x and 80 minus x. h equals x times tan 60°, so h equals x√3. Also, tan 30° equals h divided by 80 minus x. So, 1 divided by √3 equals x√3 divided by 80 minus x. 80 minus x equals 3x. 4x equals 80, so x equals 20 m. Other distance equals 60 m. Height h equals 20√3 m.
Question 11: A TV tower stands on a canal bank. From the opposite bank, elevation is 60°. From a point 20 m back, elevation is 30°, as in Figure 9.12. Find tower height and canal width. Solution: Let width be x. h equals x times tan 60°, so h equals x√3. From the second point, tan 30° equals h divided by x plus 20. So, 1 divided by √3 equals x√3 divided by x plus 20. x plus 20 equals 3x. 2x equals 20, so x equals 10 m. Height h equals 10√3 m.
Question 12: From a 7 m building, elevation to a cable tower top is 60°, and depression to its foot is 45°. Determine the tower height. Solution: Horizontal distance d equals 7 divided by tan 45°, which is 7 m. Upper part of tower above building equals 7 times tan 60°, which is 7√3 m. Total tower height equals 7√3 plus 7, or 7(√3 plus 1) m.
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Question 13: From a 75 m lighthouse, depression angles of two ships are 30° and 45°. One ship is behind the other. Find the distance between them. Solution: Distance to closer ship at 45° is d1 equals 75 divided by tan 45°, which is 75 m. Distance to farther ship at 30° is d2 equals 75 divided by tan 30°, which is 75√3 m. Distance between ships equals d2 minus d1, which is 75√3 minus 75, or 75(√3 minus 1) m.
Question 14: A 1.2 m tall girl spots a balloon moving horizontally at 88.2 m height. Elevation drops from 60° to 30°, as in Figure 9.13. Find distance travelled. Solution: Effective height above her eyes is 88.2 minus 1.2, which is 87 m. Initial distance equals 87 divided by tan 60°, which is 87 divided by √3. Final distance equals 87 divided by tan 30°, which is 87√3. Distance travelled equals 87√3 minus 87 divided by √3, which simplifies to 174 divided by √3, or 58√3 m.
Question 15: A car approaches a tower. Depression changes from 30° to 60° in 6 seconds. Find time to reach the foot from the second point. Solution: Let tower height be h. Initial distance from foot equals h divided by tan 30°, which is h√3. Final distance from foot equals h divided by tan 60°, which is h divided by √3. Distance covered in 6 seconds equals h√3 minus h divided by √3, which is 2h divided by √3. Speed equals distance divided by time, so speed is (2h divided by √3) divided by 6, which simplifies to h divided by 3√3. Remaining distance to cover is h divided by √3. Time equals remaining distance divided by speed, which is (h divided by √3) divided by (h divided by 3√3), giving exactly 3 seconds.
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Finally, let us review the chapter summary. First, the line of sight is the line drawn from the eye of an observer to the point in the object viewed. Second, the angle of elevation is the angle formed by the line of sight with the horizontal when the object is above the horizontal level. Third, the angle of depression is the angle formed by the line of sight with the horizontal when the object is below the horizontal level. Finally, the height, length, or distance between objects can be accurately determined using trigonometric ratios. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]