Some Applications Of Trigonometry - Class 10 Maths NCERT

Bringing Trigonometry to Life: The Real World Connection

In your previous studies, you learned the fundamental trigonometric ratios – sine, cosine, and tangent – and how they relate the angles of a right-angled triangle to the lengths of its sides. The chapter 'Some Applications of Trigonometry' builds directly upon this knowledge, showing you how these ratios can be practically applied to solve real-world problems involving heights and distances. The core idea revolves around forming imaginary right-angled triangles in various scenarios. When you look up at an object, say the top of a building or a bird in the sky, the line from your eye to the object is called the line of sight. If the object is above your horizontal eye level, the angle formed between your line of sight and the horizontal line passing through your eye is known as the angle of elevation. Conversely, when you look down at an object, like a boat in the sea from a lighthouse or a car on the road from a balcony, the line of sight again connects your eye to the object. If the object is below your horizontal eye level, the angle formed between your line of sight and the horizontal line is termed the angle of depression. It's crucial to remember that the angle of elevation of an object from point A is equal to the angle of depression of point A from the object, provided both are measured from the horizontal. These angles, along with known distances or heights, allow us to set up right-angled triangles. By identifying the known side (opposite, adjacent, or hypotenuse) and the unknown side we need to find, we can select the appropriate trigonometric ratio (SOH CAH TOA) to formulate an equation and solve for the unknown height or distance. This systematic approach transforms complex real-world measurement challenges into straightforward mathematical problems.

Key Terms You Must Know

Line of Sight

The straight line joining the eye of an observer to the object being viewed.

Horizontal Level

A line parallel to the ground, passing through the eye of the observer.

Angle of Elevation

When an observer looks at an object above their horizontal level, the angle formed between the line of sight and the horizontal line is called the angle of elevation.

Angle of Depression

When an observer looks at an object below their horizontal level, the angle formed between the line of sight and the horizontal line is called the angle of depression.

Your Step-by-Step Guide to Solving Problems

1. Read and Understand the Problem — Carefully read the problem statement. Identify what is given and what needs to be found. Pay attention to keywords like 'angle of elevation' or 'angle of depression'.
2. Draw a Neat and Labelled Diagram — This is the most crucial step. Represent the ground, objects (like towers, buildings, ships), and the observer with a clear, proportionate sketch. Mark the line of sight, horizontal level, and the given angles of elevation or depression. Always ensure you are drawing right-angled triangles.
3. Identify Right-Angled Triangles and Knowns/Unknowns — Locate all right-angled triangles in your diagram. Label the known values (heights, distances, angles) and assign variables (like 'x' or 'h') to the unknown quantities you need to calculate.
4. Choose the Correct Trigonometric Ratio — Based on the angle and the sides involved (opposite, adjacent, hypotenuse) in your right-angled triangle, select the appropriate trigonometric ratio (sin, cos, or tan) that connects the known and unknown values. Remember SOH CAH TOA!
5. Formulate and Solve the Equation — Set up the trigonometric equation using the chosen ratio. Substitute the known values and then solve the equation for the unknown variable. Use the standard values of trigonometric ratios for specific angles (0°, 30°, 45°, 60°, 90°).
6. State the Final Answer with Units — Once you've found the numerical value, write down your answer clearly, including the appropriate units (e.g., meters, centimeters). Review your answer to ensure it makes sense in the context of the problem.

Solved Examples to Build Your Confidence

• Example 1: Finding the Height of a Tower 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: Step 1: Draw a diagram. Let AB be the tower and C be the point on the ground. We have a right-angled triangle ABC, with the right angle at B. `` A (Top of Tower) | | h (Height) | | B-----15m-----C (Foot of Tower) (Point on Ground) `` Step 2: Identify knowns and unknowns. - Distance from foot of tower (BC) = 15 m - Angle of elevation (∠ACB) = 60° - Height of tower (AB) = h (unknown) Step 3: Choose the trigonometric ratio. - We know the adjacent side (BC) and need to find the opposite side (AB) relative to the angle 60°. - The ratio connecting opposite and adjacent is tangent: tanθ = Opposite/Adjacent. Step 4: Formulate and solve the equation. - tan(60°) = AB/BC - √3 = h/15 - h = 15√3 m Final answer: The height of the tower is 15√3 meters.
• Example 2: Angle of Depression from a Lighthouse A 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°. What is the height of the chimney? Solution: Step 1: Draw a diagram. Let AB be the chimney and CD be the observer. Draw a line DE parallel to BC from the observer's eye (D) to the chimney (AE). This forms a right-angled triangle ADE. `` A (Top of Chimney) | | E----- | \ | \ (Line of Sight) h' D (Observer's Eye) | | | | B------C (Ground) (Observer's Foot) `` Step 2: Identify knowns and unknowns. - Observer's height (CD) = 1.5 m - Distance from chimney (BC = DE) = 28.5 m - Angle of elevation (∠ADE) = 45° - Height of chimney (AB) = AE + EB = AE + CD - Let AE = h' Step 3: Choose the trigonometric ratio. - In ΔADE, we know the adjacent side (DE) and need the opposite side (AE). - Use tanθ = Opposite/Adjacent. Step 4: Formulate and solve the equation. - tan(45°) = AE/DE - 1 = h'/28.5 - h' = 28.5 m Step 5: Calculate total height. - Height of chimney AB = AE + EB = h' + CD = 28.5 + 1.5 = 30 m Final answer: The height of the chimney is 30 meters.
• Example 3: Observer Moving Closer to a Building 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. Solution: Step 1: Draw a diagram. Let BC be the building (20 m) and AB be the transmission tower. Let D be the point on the ground. We have two right-angled triangles, ΔDBC and ΔDAC. `` A (Top of Tower) | h (Tower Height) | B (Bottom of Tower / Top of Building) | 20m (Building Height) | C----------------D (Foot of Building) (Point on Ground) `` Step 2: Identify knowns and unknowns. - Height of building (BC) = 20 m - Angle of elevation to bottom of tower (∠BDC) = 45° - Angle of elevation to top of tower (∠ADC) = 60° - Height of tower (AB) = h (unknown) - Let CD = x (distance from building, unknown) - AC = AB + BC = h + 20 m Step 3: Use trigonometric ratios in both triangles. - In ΔDBC: - tan(45°) = BC/CD - 1 = 20/x - x = 20 m - In ΔDAC: - tan(60°) = AC/CD - √3 = (h + 20)/x - Substitute x = 20: - √3 = (h + 20)/20 - 20√3 = h + 20 Step 4: Solve for h. - h = 20√3 - 20 - h = 20(√3 - 1) - Using √3 ≈ 1.732: - h = 20(1.732 - 1) - h = 20(0.732) - h = 14.64 m Final answer: The height of the transmission tower is 20(√3 - 1) meters or approximately 14.64 meters.

YoLearn.ai Exam Tips for Success!

To ace questions from 'Some Applications of Trigonometry', follow these smart strategies:

1. Diagram is Key: Always draw a clear, large, and accurately labeled diagram. Incorrect diagrams lead to incorrect solutions. Mark all given values and unknown variables clearly.
2. Understand Angles: Be absolutely clear about the difference between the angle of elevation and the angle of depression. Remember that the angle of depression from an object to a point is equal to the angle of elevation from the point to the object (alternate interior angles).
3. Correct Ratio: Choose the appropriate trigonometric ratio (sin, cos, or tan) based on the sides you know and the side you need to find relative to the given angle. A common mistake is using sine when tangent is required.
4. Standard Values: Memorize the trigonometric values for special angles (0°, 30°, 45°, 60°, 90°). These are frequently used and not having them handy can slow you down or lead to errors.
5. Step-by-Step: Show all your steps clearly, from drawing the diagram to writing the equations and performing calculations. This helps in error detection and fetches marks even if the final answer is slightly off.

Practice Questions with Solutions

• Q: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. A: Step 1: Draw a diagram. Let AB be the pole and AC be the rope. Triangle ABC is a right-angled triangle at B. Step 2: Identify knowns and unknowns. Length of rope (hypotenuse AC) = 20 m. Angle (∠ACB) = 30°. Height of pole (opposite side AB) = h (unknown). Step 3: Choose the trigonometric ratio. We have hypotenuse and need opposite side. So, use sine: sinθ = Opposite/Hypotenuse. Step 4: Formulate and solve. sin(30°) = AB/AC => 1/2 = h/20 => h = 20/2 = 10 m. Final answer: The height of the pole is 10 meters.
• Q: The angle of elevation of the top of a building from the foot of the tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. A: Step 1: Draw a diagram. Let AB be the building (height h) and CD be the tower (height 50m). Let the distance between them be BC = x. We have two right-angled triangles, ΔABC and ΔDCB. Step 2: Identify knowns and unknowns. - In ΔDCB: CD = 50 m, ∠DBC = 60°, BC = x. - In ΔABC: AB = h, ∠ACB = 30°, BC = x. Step 3: Solve for x using ΔDCB. tan(60°) = CD/BC => √3 = 50/x => x = 50/√3 m. Step 4: Solve for h using ΔABC and the value of x. tan(30°) = AB/BC => 1/√3 = h/x => 1/√3 = h / (50/√3) => h = (1/√3) * (50/√3) = 50/3 m. Final answer: The height of the building is 50/3 meters.
• Q: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. A: Step 1: Draw a diagram. Let AB be the building (30 m). Let P be the initial position of the boy and Q be his final position. His height is 1.5 m. Draw a horizontal line PR from his eyes parallel to the ground. Let R be a point on the building. So, AR = 30 - 1.5 = 28.5 m. Step 2: Identify knowns and unknowns. - Height of building above eye level (AR) = 28.5 m. - Initial angle (∠APR) = 30°. Final angle (∠AQR) = 60°. - Let PR = y and QR = x. Distance walked = PQ = y - x. Step 3: In ΔAQR: tan(60°) = AR/QR => √3 = 28.5/x => x = 28.5/√3. Step 4: In ΔAPR: tan(30°) = AR/PR => 1/√3 = 28.5/y => y = 28.5√3. Step 5: Calculate distance walked. PQ = y - x = 28.5√3 - 28.5/√3 = 28.5(√3 - 1/√3) = 28.5 ( (3-1)/√3 ) = 28.5 (2/√3) = 57/√3 = 57√3/3 = 19√3 m. Final answer: The boy walked 19√3 meters towards the building.
• Q: An aeroplane at an altitude of 1200 m finds that two ships are sailing towards it in the same direction. The angles of depression of the ships from the aeroplane are observed to be 60° and 30° respectively. Find the distance between the two ships. A: Step 1: Draw a diagram. Let A be the position of the aeroplane, and B be the point directly below it on the ground. Let C and D be the positions of the two ships. AB = 1200 m. We have two right-angled triangles, ΔABC and ΔABD. Step 2: Identify knowns and unknowns. - Height of aeroplane (AB) = 1200 m. - Angles of depression: ∠XAC = 60° (so ∠ACB = 60° by alternate interior angles), ∠XAD = 30° (so ∠ADB = 30°). - Let BC = x and BD = y. Distance between ships = CD = y - x. Step 3: In ΔABC: tan(60°) = AB/BC => √3 = 1200/x => x = 1200/√3 = 400√3 m. Step 4: In ΔABD: tan(30°) = AB/BD => 1/√3 = 1200/y => y = 1200√3 m. Step 5: Calculate distance between ships. CD = y - x = 1200√3 - 400√3 = 800√3 m. Final answer: The distance between the two ships is 800√3 meters.

Frequently Asked Questions