Worksheet
Explore real-world applications of trigonometry in measuring heights, distances, and angles in various fields such as astronomy, navigation, and architecture.
Some Applications of Trigonometry - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
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
Strengthen your understanding with fundamental questions about the chapter.
Questions
Explain the concept of angle of elevation and angle of depression with real-life examples.
Think about how you look at objects above or below your eye level and the angle your line of sight makes with the horizontal.
Solution
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.
Use the tangent ratio for the angle of elevation to relate the height of the tower and the distance from the point to the tower.
Solution
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.
Consider the ladder as the hypotenuse of a right triangle and use trigonometric ratios to find the unknown sides.
Solution
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.
Use the tangent of the angles of elevation to set up equations relating the heights and the distance from the point to the building.
Solution
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.
Set up equations using the tangent of the sun's altitude angles and solve for the height of the tower.
Solution
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.
Use the sine ratio for the angle of inclination to relate the height of the kite and the length of the string.
Solution
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.
Use the tangent of the angles of elevation and depression to find the height of the tower relative to the building.
Solution
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.
Set up equations using the tangent of the angles of elevation and solve for the height and distances.
Solution
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.
Use the tangent of the angles of elevation from both points to set up equations and solve for the height and width.
Solution
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.
Use the tangent of the angles of depression to relate the height of the tower and the distances, and consider the uniform speed of the car to find the time.
Solution
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.
Some Applications of Trigonometry - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Some Applications of Trigonometry to prepare for higher-weightage questions in Class X.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
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.
Remember that tan(θ) = opposite/adjacent in a right-angled triangle.
Solution
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.
Break the problem into finding the ladder's length first, then its horizontal distance from the pole.
Solution
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.
First find the distance to the building using the angle of elevation to its bottom, then use that to find the tower's height.
Solution
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.
Use the angle to the pedestal to express distance in terms of h, then apply the angle to the statue.
Solution
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.
Set up equations for each angle of elevation and solve the system for x and h.
Solution
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.
Use the angle of depression to find the distance to the tower, then the angle of elevation to find its height.
Solution
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.
Calculate horizontal distances at both angles and find the difference.
Solution
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.
Express distances in terms of h, find speed, then time to cover the remaining distance.
Solution
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.
Focus on the direction of the line of sight relative to the horizontal.
Solution
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.
Think about forming a right triangle with the object and using known angle and distance.
Solution
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.
Some Applications of Trigonometry - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Some Applications of Trigonometry in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
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.
Consider which trigonometric ratio relates the angle of elevation to the opposite side (height) and the adjacent side (distance from the tower).
Solution
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.
Break the problem into finding the ladder's length first, then use another trigonometric ratio to find the distance from the pole.
Solution
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.
First find the distance using the building's height and the angle of elevation to its top, then use the 45° angle to find the total height including the flagstaff.
Solution
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.
Set up two equations based on the two angles and solve for the unknowns.
Solution
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.
Use the angles of depression to relate the heights and distances, remembering that the angle of depression equals the angle of elevation from the lower point.
Solution
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.
Calculate the distances to each bank separately and add them together to get the total width.
Solution
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 rope acts as the hypotenuse of a right-angled triangle, with the pole as the opposite side to the angle.
Solution
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 broken part forms a right-angled triangle with the ground and the remaining part of the tree.
Solution
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 slide length is the hypotenuse of a right-angled triangle where the height is the opposite side to the angle of inclination.
Solution
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.
Use the sine of the angle to relate the height to the string length.
Solution
The string length is the hypotenuse when the height is the opposite side to the angle. Thus, length = 60 / sin 60° ≈ 69.28 m.
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Explore the basics of trigonometry, including angles, triangles, and the fundamental trigonometric ratios: sine, cosine, and tangent.
