Multi-Step Trigonometry Problems
Multi-step trigonometry problems involve two or more angles of elevation and/or depression, moving objects, or shadow-based calculations. These are advanced applications of trigonometric ratios covered in Class 10 NCERT under "Some Applications of Trigonometry".
These problems typically require setting up two right triangles that share a common side, then using simultaneous equations to find the unknown height or distance.
Board examinations frequently include 4-mark or 5-mark questions of this type. A systematic approach — drawing the figure, labelling all known and unknown quantities, and identifying the right triangles — is essential for success.
What is Multi-Step Trigonometry Problems?
Definition: A multi-step trigonometry problem is one where a single application of a trigonometric ratio in one right triangle is not sufficient. The problem requires:
- Two or more right triangles sharing a common side (usually the height).
- Two angles of elevation/depression from different positions.
- Elimination of one unknown to find the required quantity.
Key terms:
- Angle of elevation: The angle formed above the horizontal when looking up at an object.
- Angle of depression: The angle formed below the horizontal when looking down at an object.
- Line of sight: The straight line from the observer's eye to the object.
Multi-Step Trigonometry Problems Formula
Trigonometric Ratios Used:
tan θ = Opposite / Adjacent
sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
Standard values used:
- tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
- sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2
- cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2
Types and Properties
Multi-step trigonometry problems can be classified as:
- Two-position problems: Observer moves towards or away from an object; angles of elevation from both positions are given.
- Two-angle-from-top problems: From the top of a building, angles of depression to two objects on the ground are given.
- Moving object problems: A boat/car moves in a straight line; angles change as it moves.
- Shadow problems: Length of shadow changes with the angle of the sun.
- Height-on-height problems: A flagstaff on top of a building; find the height of the flagstaff.
- Between-two-buildings problems: Two buildings of different heights; angles from one roof to the other.
Solved Examples
Example 1: Two Positions — Moving Towards a Tower
Problem: The angles of elevation of the top of a tower from two points at distances a and b from the base (a > b), in the same straight line and on the same side, are 30° and 60° respectively. Prove that the height of the tower is √(ab).
Solution:
Given:
- Let the height of the tower = h
- Distance of first point from base = a (angle = 30°)
- Distance of second point from base = b (angle = 60°)
From the first triangle:
- tan 30° = h/a
- 1/√3 = h/a
- h = a/√3 ... (i)
From the second triangle:
- tan 60° = h/b
- √3 = h/b
- h = b√3 ... (ii)
Multiply (i) and (ii):
- h × h = (a/√3)(b√3)
- h² = ab
- h = √(ab)
Answer: The height of the tower is √(ab). Hence proved.
Example 2: Flagstaff on a Building
Problem: From a point on the ground, the angles of elevation of the bottom and top of a flagstaff fixed on the top of a 20 m high building are 45° and 60° respectively. Find the height of the flagstaff. (Use √3 = 1.732)
Solution:
Given:
- Height of building = 20 m
- Let height of flagstaff = h m
- Let distance of point from building = x m
From triangle with building:
- tan 45° = 20/x → 1 = 20/x → x = 20 m
From triangle with building + flagstaff:
- tan 60° = (20 + h)/x
- √3 = (20 + h)/20
- 20√3 = 20 + h
- h = 20√3 − 20 = 20(√3 − 1)
- h = 20(1.732 − 1) = 20 × 0.732 = 14.64 m
Answer: The height of the flagstaff is 14.64 m.
Example 3: Angles of Depression from a Cliff
Problem: From the top of a 75 m high cliff, the angles of depression of two ships are 30° and 45°. If the ships are on the same side and in line with the base of the cliff, find the distance between the two ships. (Use √3 = 1.732)
Solution:
Given:
- Height of cliff = 75 m
- Angle of depression to nearer ship = 45°
- Angle of depression to farther ship = 30°
For the nearer ship:
- tan 45° = 75/d₁ → d₁ = 75 m
For the farther ship:
- tan 30° = 75/d₂ → 1/√3 = 75/d₂ → d₂ = 75√3 m
Distance between ships:
- = d₂ − d₁ = 75√3 − 75 = 75(√3 − 1)
- = 75(1.732 − 1) = 75 × 0.732 = 54.9 m
Answer: The distance between the two ships is 54.9 m.
Example 4: Shadow Length Problem
Problem: 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. (Use √3 = 1.732)
Solution:
Given:
- Let height of tower = h m
- Let shadow length when angle = 60° be x m
- Shadow length when angle = 30° = (x + 40) m
When sun's altitude = 60°:
- tan 60° = h/x → √3 = h/x → x = h/√3 ... (i)
When sun's altitude = 30°:
- tan 30° = h/(x + 40) → 1/√3 = h/(x + 40) → x + 40 = h√3 ... (ii)
Substitute (i) in (ii):
- h/√3 + 40 = h√3
- 40 = h√3 − h/√3 = h(√3 − 1/√3) = h(3 − 1)/√3 = 2h/√3
- h = 40√3/2 = 20√3
- h = 20 × 1.732 = 34.64 m
Answer: The height of the tower is 34.64 m.
Example 5: Two Buildings Problem
Problem: Two buildings are on opposite sides of a road 15 m wide. From the top of the first building (12 m high), the angle of elevation of the top of the second building is 60° and the angle of depression of its base is 45°. Find the height of the second building.
Solution:
Given:
- Width of road = 15 m, Height of first building = 12 m
- Angle of depression to base of second building = 45°
- Angle of elevation to top of second building = 60°
Using angle of depression (to verify width):
- tan 45° = 12/width → width = 12 m
But width is given as 15 m. So using tan 45° for verification shows the horizontal distance from top of first to base of second = 15 m (the road width, since buildings are on opposite sides).
Let excess height of second building above first = h
- tan 60° = h/15
- √3 = h/15
- h = 15√3 = 15 × 1.732 = 25.98 m
Total height of second building:
- = 12 + 15√3 = 12 + 25.98 = 37.98 m
Answer: The height of the second building is 37.98 m (or 12 + 15√3 m).
Example 6: Moving Boat Problem
Problem: From the top of a 100 m high lighthouse, the angles of depression of two boats on opposite sides are 30° and 45°. Find the distance between the boats. (Use √3 = 1.732)
Solution:
Given:
- Height of lighthouse = 100 m
- Angle of depression to boat A = 30°
- Angle of depression to boat B = 45°
- Boats are on opposite sides
Distance of boat A:
- tan 30° = 100/d₁ → d₁ = 100√3 m
Distance of boat B:
- tan 45° = 100/d₂ → d₂ = 100 m
Since boats are on opposite sides:
- Total distance = d₁ + d₂ = 100√3 + 100 = 100(√3 + 1)
- = 100(1.732 + 1) = 100 × 2.732 = 273.2 m
Answer: The distance between the two boats is 273.2 m.
Example 7: Cloud and Reflection Problem
Problem: The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud above the lake.
Solution:
Given:
- Height of observation point above lake = 60 m
- Let height of cloud above lake = h m
- Height of cloud above observer = (h − 60) m
- Depth of reflection below observer = (h + 60) m
- Let horizontal distance = x m
Using angle of elevation (30°):
- tan 30° = (h − 60)/x
- 1/√3 = (h − 60)/x → x = (h − 60)√3 ... (i)
Using angle of depression (60°):
- tan 60° = (h + 60)/x
- √3 = (h + 60)/x → x = (h + 60)/√3 ... (ii)
Equating (i) and (ii):
- (h − 60)√3 = (h + 60)/√3
- 3(h − 60) = h + 60
- 3h − 180 = h + 60
- 2h = 240
- h = 120 m
Answer: The height of the cloud above the lake is 120 m.
Example 8: Pilot Observing Two Ships
Problem: An aeroplane at an altitude of 200 m observes the angles of depression of two ships in the same direction to be 45° and 30°. Find the distance between the ships.
Solution:
Given:
- Altitude = 200 m
- Angle of depression to nearer ship = 45°
- Angle of depression to farther ship = 30°
Nearer ship:
- tan 45° = 200/d₁ → d₁ = 200 m
Farther ship:
- tan 30° = 200/d₂ → d₂ = 200√3 m
Distance between ships:
- = d₂ − d₁ = 200√3 − 200 = 200(√3 − 1)
- = 200 × 0.732 = 146.4 m
Answer: The distance between the two ships is 146.4 m.
Real-World Applications
Multi-step trigonometry problems apply directly in:
- Surveying: Measuring heights of mountains, buildings, and towers using theodolites from two positions.
- Navigation: Pilots and ship captains use angles to calculate distances to landmarks.
- Defence: Finding the distance of an enemy target using two observation posts.
- Architecture: Determining the height of structures when direct measurement is not possible.
- Astronomy: Measuring distances to celestial objects using parallax (two observations from different positions).
Key Points to Remember
- Always draw a clear, labelled figure before writing equations.
- The angle of depression from A to B equals the angle of elevation from B to A (alternate interior angles).
- Most multi-step problems use tan θ because height (opposite) and distance (adjacent) are the unknowns.
- When two angles are given from the same point, set up two equations with two unknowns and solve simultaneously.
- For objects on the same side, subtract distances; for opposite sides, add distances.
- Rationalise the denominator when the answer contains √3 or √2.
- Always use the exact trigonometric values for 30°, 45°, and 60° — do not use decimal approximations during calculation.
- Substitute √3 = 1.732 only at the final step for numerical answers.
- Cloud-reflection problems use (h − observer height) for elevation and (h + observer height) for the reflection.
- Verify that your answer is dimensionally consistent and physically reasonable.
Practice Problems
- From the top of a 60 m high building, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower.
- The angle of elevation of the top of a tower from a point A is 45°. On walking 20 m towards the tower, the angle becomes 60°. Find the height of the tower.
- Two poles of heights 6 m and 11 m stand on a level ground. The distance between their feet is 12 m. Find the distance between their tops.
- From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the cable tower.
- A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he moves 40 m away from the bank, the angle becomes 30°. Find the width of the river.
- A straight highway leads to the foot of a tower. A man standing at the top observes a car at an angle of depression 30° which is approaching the tower at uniform speed. After 6 seconds, the angle of depression becomes 60°. Find the time taken by the car to reach the tower.
Frequently Asked Questions
Q1. What makes a trigonometry problem 'multi-step'?
A problem is multi-step when it involves two or more right triangles, requiring two or more trigonometric equations to be solved simultaneously. This typically happens when two angles are given from different positions or to different objects.
Q2. Which trigonometric ratio is used most often in height and distance problems?
The tangent ratio (tan θ = perpendicular/base) is used most often because height and distance problems typically involve the vertical height (perpendicular) and horizontal distance (base) of a right triangle.
Q3. How do you handle problems with objects on opposite sides?
When two objects are on opposite sides of a tower or lighthouse, the total distance between them is the sum of their individual distances from the base. When on the same side, the distance between them is the difference.
Q4. What is the cloud-reflection method?
When a cloud is observed from a point above a lake, the height of the cloud above the observer is (h − d) where h is the cloud's height above the lake and d is the observer's height. The depth of the reflection below the observer is (h + d). Two equations are formed and solved simultaneously.
Q5. How do you find the speed of a moving object using trigonometry?
Observe the object from a fixed point at two different times. Calculate the distance covered between the two observations using trigonometric ratios. Then speed = distance/time.
Q6. Why must we rationalise the denominator?
Rationalising the denominator (removing surds from the denominator) gives a cleaner form and is the standard mathematical convention. Multiply numerator and denominator by the conjugate or the surd to rationalise.
