This chapter explores how trigonometry is applied in real-life situations, particularly in measuring heights and distances.
Some Applications of Trigonometry - Practice Worksheet
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This worksheet covers essential long-answer questions to help you build confidence in Some Applications of Trigonometry from Mathematics for Class X (Mathematics).
Basic comprehension exercises
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Questions
Explain the concept of angle of elevation and angle of depression with real-life examples.
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. For example, when you look up at the top of a building from the ground, the angle your line of sight makes with the horizontal is the angle of elevation. Conversely, the angle of depression is the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level. An example is when you look down from a balcony at an object on the ground. These concepts are crucial in trigonometry for solving problems related to heights and distances without direct measurement.
A tower stands vertically on the ground. From a point on the ground, 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.
To find the height of the tower, we can use the tangent of the angle of elevation. In this scenario, the distance from the point to the foot of the tower is the adjacent side (15 m), and the height of the tower is the opposite side to the angle of elevation. Using the formula tan(θ) = opposite/adjacent, we have tan(60°) = height/15. Since tan(60°) = √3, the height = 15 * √3 ≈ 25.98 m. Therefore, the height of the tower is approximately 25.98 meters.
An electrician needs to reach a point 1.3 m below the top of a 5 m tall pole to repair an electric fault. She uses a ladder inclined at an angle of 60° to the horizontal. Find the length of the ladder and how far from the foot of the pole she should place the foot of the ladder.
The electrician needs to reach a point 3.7 m above the ground (5 m - 1.3 m). The ladder forms a right triangle with the ground and the pole. To find the length of the ladder (hypotenuse), we use the sine of the angle of inclination: sin(60°) = opposite/hypotenuse = 3.7/length of ladder. Solving gives the length of the ladder ≈ 4.28 m. To find how far from the pole the ladder's foot is placed (adjacent side), we use the cosine of the angle: cos(60°) = adjacent/hypotenuse = distance/4.28. Solving gives the distance ≈ 2.14 m.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed on a 20 m high building are 45° and 60° respectively. Find the height of the transmission tower.
Let's denote the height of the transmission tower as 'h'. The total height from the ground to the top of the tower is 20 m + h. For the angle of elevation of the bottom of the tower (45°), tan(45°) = 20/distance ⇒ distance = 20 m. For the angle of elevation of the top of the tower (60°), tan(60°) = (20 + h)/20 ⇒ √3 = (20 + h)/20 ⇒ h = 20(√3 - 1) ≈ 14.64 m. Therefore, the height of the transmission tower is approximately 14.64 meters.
The shadow of a tower standing on 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.
Let the height of the tower be 'h' and the length of the shadow when the sun's altitude is 60° be 'x'. When the sun's altitude is 30°, the shadow length becomes x + 40. Using the tangent ratio for both angles, we have tan(60°) = h/x ⇒ h = x√3, and tan(30°) = h/(x + 40) ⇒ h = (x + 40)/√3. Setting the two expressions for h equal gives x√3 = (x + 40)/√3 ⇒ 3x = x + 40 ⇒ x = 20. Therefore, h = 20√3 ≈ 34.64 m. The height of the tower is approximately 34.64 meters.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming there is no slack in the string.
The situation forms a right triangle with the height of the kite as the opposite side (60 m), the length of the string as the hypotenuse, and the angle of inclination as 60°. Using the sine ratio: sin(60°) = opposite/hypotenuse = 60/length of string ⇒ length of string = 60/sin(60°) = 60/(√3/2) ≈ 69.28 m. Therefore, the length of the string is approximately 69.28 meters.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let the height of the cable tower be 'h' and the distance between the buildings be 'd'. From the top of the building, the angle of depression to the foot of the tower is 45°, so tan(45°) = 7/d ⇒ d = 7 m. The angle of elevation to the top of the tower is 60°, so tan(60°) = (h - 7)/d ⇒ √3 = (h - 7)/7 ⇒ h = 7√3 + 7 ≈ 19.12 m. Therefore, the height of the cable tower is approximately 19.12 meters.
Two poles of equal heights are standing opposite each other on either side of a road 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°. Find the height of the poles and the distances of the point from the poles.
Let the height of each pole be 'h' and the distances from the point to the two poles be 'x' and '80 - x'. Using the tangent ratio for both angles, we have tan(60°) = h/x ⇒ h = x√3, and tan(30°) = h/(80 - x) ⇒ h = (80 - x)/√3. Setting the two expressions for h equal gives x√3 = (80 - x)/√3 ⇒ 3x = 80 - x ⇒ x = 20. Therefore, h = 20√3 ≈ 34.64 m. The distances from the point to the poles are 20 m and 60 m. The height of the poles is approximately 34.64 meters.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Let the height of the tower be 'h' and the width of the canal be 'd'. From the first point, tan(60°) = h/d ⇒ h = d√3. From the second point, tan(30°) = h/(d + 20) ⇒ h = (d + 20)/√3. Setting the two expressions for h equal gives d√3 = (d + 20)/√3 ⇒ 3d = d + 20 ⇒ d = 10. Therefore, h = 10√3 ≈ 17.32 m. The height of the tower is approximately 17.32 meters, and the width of the canal is 10 meters.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Let the height of the tower be 'h' and the initial distance of the car from the tower be 'd'. The angle of depression changes from 30° to 60° as the car approaches. Using the tangent ratio, tan(30°) = h/d ⇒ d = h√3, and tan(60°) = h/(d - distance covered in 6 seconds). Let the speed of the car be 'v', so the distance covered in 6 seconds is 6v. Therefore, tan(60°) = h/(h√3 - 6v) ⇒ √3 = h/(h√3 - 6v) ⇒ h = 3h - 6v√3 ⇒ 2h = 6v√3 ⇒ h = 3v√3. The remaining distance when the angle is 60° is h/√3 = 3v. The time to cover this distance at speed 'v' is 3v/v = 3 seconds. Therefore, the car takes 3 more seconds to reach the foot of the tower.
Question 1 of 10
Explain the concept of angle of elevation and angle of depression with real-life examples.
Some Applications of Trigonometry - Mastery Worksheet
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Intermediate analysis exercises
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Questions
A tower stands vertically on the ground. From a point on the ground, 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.
To find the height of the tower, we use the tangent of the angle of elevation. tan(60°) = height of the tower / distance from the point to the tower. Thus, height = 15 * tan(60°) = 15 * √3 ≈ 25.98 m.
An electrician needs to reach a point 1.3 m below the top of a 5 m tall pole to repair an electric fault. She uses a ladder inclined at an angle of 60° to the horizontal. Find the length of the ladder and how far from the foot of the pole she should place the ladder.
The height to be reached is 5 - 1.3 = 3.7 m. Using sin(60°) = opposite/hypotenuse, the length of the ladder (hypotenuse) = 3.7 / sin(60°) ≈ 4.28 m. The distance from the pole is found using cos(60°) = adjacent/hypotenuse, so distance = 4.28 * cos(60°) ≈ 2.14 m.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Let the height of the tower be h. The total height from the ground is 20 + h. Using tan(45°) = 1 = 20 / distance, so distance = 20 m. Then, tan(60°) = √3 = (20 + h) / 20. Solving gives h = 20(√3 - 1) ≈ 14.64 m.
A statue stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and of the top of the pedestal is 45°. If the statue is 1.6 m tall, find the height of the pedestal.
Let the pedestal's height be h. The total height is h + 1.6. Using tan(45°) = 1 = h / distance, so distance = h. Then, tan(60°) = √3 = (h + 1.6) / h. Solving gives h = 1.6 / (√3 - 1) ≈ 2.19 m.
Two poles of equal heights stand opposite each other on either side of an 80 m wide road. From a point between them on the road, the angles of elevation of the tops are 60° and 30°. Find the height of the poles and the distances of the point from the poles.
Let the height be h and distances be x and 80 - x. Then, tan(60°) = √3 = h / x and tan(30°) = 1/√3 = h / (80 - x). Solving gives h = x√3 and h = (80 - x)/√3. Equating gives x = 20 m, so h = 20√3 ≈ 34.64 m, and distances are 20 m and 60 m.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
The angle of depression of 45° means the distance to the tower is equal to the building's height, 7 m. Then, tan(60°) = √3 = (tower height - 7) / 7. Solving gives tower height = 7 + 7√3 ≈ 19.12 m.
A 1.2 m tall girl spots a balloon at an angle of elevation of 60°. After some time, the angle reduces to 30°. Find the distance travelled by the balloon if its height is 88.2 m from the ground.
Initial distance to balloon: tan(60°) = (88.2 - 1.2) / d1 ⇒ d1 = 87 / √3 ≈ 50.23 m. Final distance: tan(30°) = 87 / d2 ⇒ d2 = 87√3 ≈ 150.69 m. Distance travelled = d2 - d1 ≈ 100.46 m.
A man observes a car at an angle of depression of 30°, which is approaching the foot of a tower. Six seconds later, the angle is 60°. Find the time taken by the car to reach the foot of the tower from the second observation.
Let the tower's height be h. Initial distance: tan(30°) = h / d1 ⇒ d1 = h√3. Final distance: tan(60°) = h / d2 ⇒ d2 = h / √3. Distance covered in 6 seconds: d1 - d2 = h(√3 - 1/√3) = 2h/√3. Speed = distance / time = (2h/√3) / 6 = h/(3√3). Time to cover d2: d2 / speed = (h/√3) / (h/(3√3)) = 3 seconds.
Compare and contrast the concepts of angle of elevation and angle of depression with examples.
Angle of elevation is the angle between the line of sight and the horizontal when looking upwards, e.g., looking at a tower's top. Angle of depression is when looking downwards, e.g., looking at a boat from a cliff. Both involve the horizontal but differ in the direction of the line of sight.
Explain how trigonometric ratios can be used to determine the height of an object without directly measuring it, using an example.
Trigonometric ratios relate angles to side ratios in right triangles. For example, to find a tree's height, measure the distance from the tree and the angle of elevation to its top. Using tan(angle) = height / distance, solve for height. This method is useful for inaccessible objects.
Question 1 of 10
A tower stands vertically on the ground. From a point on the ground, 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.
Some Applications of Trigonometry - Challenge Worksheet
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Advanced critical thinking
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Questions
A tower stands vertically on the ground. From a point on the ground, 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. Justify your method of calculation.
Using the tangent of the angle of elevation, tan 60° = height / distance from the tower. Thus, height = 15 * √3 ≈ 25.98 m. This method is chosen because it directly relates the known distance and angle to the unknown height.
An electrician needs to reach a point 1.3 m below the top of a 5 m tall pole to repair an electric fault. She uses a ladder inclined at 60° to the horizontal. Calculate the length of the ladder and how far from the pole she should place its foot.
The height to reach is 5 - 1.3 = 3.7 m. Using sin 60° = opposite / hypotenuse, the ladder length is 3.7 / (√3/2) ≈ 4.28 m. The distance from the pole is found using tan 60° = opposite / adjacent, giving 3.7 / √3 ≈ 2.14 m.
From a point on the ground, the angle of elevation of the top of a building is 30° and the angle of elevation of the top of a flagstaff on the building is 45°. The building is 10 m tall. Find the height of the flagstaff and the distance from the point to the building.
The distance to the building is 10√3 ≈ 17.32 m using tan 30°. The flagstaff height is found by subtracting the building's height from the total height when the angle is 45°, giving 10(√3 - 1) ≈ 7.32 m.
The shadow of a tower on level ground is 40 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.
Let the height be h and the shorter shadow length be x. Then h = x√3 and h = (x + 40)/√3. Solving gives x = 20 m and h = 20√3 ≈ 34.64 m.
From the top of a multi-storeyed building, the angles of depression of the top and bottom of an 8 m tall building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
The height is 4(3 + √3) ≈ 18.93 m and the distance is 4(3 + √3) ≈ 18.93 m. This is found by setting up equations based on the angles and the height difference.
A bridge is 3 m above a river. From a point on the bridge, the angles of depression of the banks on opposite sides are 30° and 45°. Find the width of the river.
The width is the sum of the distances from the point to each bank, calculated using tan of the angles of depression. Width = 3(1 + √3) ≈ 8.20 m.
A circus artist is climbing a 20 m long rope tied from the top of a vertical pole to the ground. If the angle between the rope and the ground is 30°, find the height of the pole.
The height of the pole is the rope length times sin 30°, which is 20 * 0.5 = 10 m.
A tree breaks and the top touches the ground 8 m from the base, making a 30° angle. Find the original height of the tree.
The original height is the sum of the remaining part and the part that became the hypotenuse. Using cos 30°, the remaining part is 8√3 ≈ 13.86 m, and the broken part is 16 m, totaling ≈ 29.86 m.
Two slides in a park are for different age groups. One is 1.5 m high at 30° inclination, and the other is 3 m high at 60°. Find the lengths of both slides.
The lengths are found using sin of the inclination angles. For the first slide, length = 1.5 / sin 30° = 3 m. For the second, length = 3 / sin 60° ≈ 3.46 m.
A kite is flying at 60 m height with the string making a 60° angle with the ground. Find the length of the string assuming no slack.
The string length is the hypotenuse when the height is the opposite side to the angle. Thus, length = 60 / sin 60° ≈ 69.28 m.
Question 1 of 10
A tower stands vertically on the ground. From a point on the ground, 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. Justify your method of calculation.
This chapter explores quadratic equations, highlighting their forms and significance in real-world applications.
This chapter introduces arithmetic progressions, which are sequences of numbers generated by adding a fixed value to the previous term. Understanding these patterns is crucial for solving real-life mathematical problems.
This chapter focuses on the properties of triangles, specifically their similarity and how it can be applied in various real-world contexts.
This chapter covers the concepts of coordinate geometry, including finding distances between points and dividing line segments. Understanding these concepts is essential for solving geometry problems using algebra.
This chapter focuses on the foundational concepts of trigonometry, particularly the relationships between the angles and sides of right triangles.
This chapter explores the properties of circles, particularly focusing on tangents and their relationship with radii and secants.
This chapter focuses on sectors and segments of circles, essential concepts in geometry. Understanding these helps in solving real-life problems related to areas and measurements.
This chapter explores how to find the surface areas and volumes of various solids, including combinations of basic shapes like cubes, cones, cylinders, and spheres, essential for real-world applications.
Statistics is the chapter that deals with the collection, analysis, interpretation, presentation, and organization of data.
This chapter explores the basic concepts and definitions of probability, highlighting its significance in predicting outcomes in uncertain situations.
