How to Solve: Height and Distance Problems (Trigonometry)

How to Solve: Height and Distance Problems (Trigonometry)

Introduction

"Imagine you’re an architect designing a bridge, or a pilot calculating the perfect landing angle—height and distance problems are the secret math behind these real-world decisions. Master this, and you’ll ace not just your trigonometry exam, but also questions on physics, engineering, and even navigation!

What You Need To Know First

Before tackling height and distance problems, you must already understand: 1. Basic trigonometric ratios (sine, cosine, tangent) – How to relate angles to sides in right-angled triangles. 2. Angle of elevation and depression – The difference between looking up (elevation) and looking down (depression). 3. Pythagoras’ theorem – For finding missing sides in right-angled triangles (though trigonometry is often faster).

If any of these are unclear, review them before proceeding.

Key Vocabulary

Formulas To Know

Pro Tip: Most height and distance problems use tangent (tan) because they involve height (opposite) and distance (adjacent). Only use sine or cosine if the hypotenuse is given or required.

Step-by-Step Method

Step 1: Draw a Clear Diagram

• Sketch the scenario as a right-angled triangle.
• Label:
• The observer’s position (where the angle is measured from).
• The object (e.g., a building, tree, or tower).
• The horizontal distance (ground level from observer to object).
• The vertical height (from ground to top of object).
• The angle of elevation/depression (with a clear arrow).

Why? A diagram prevents confusion about which side is opposite or adjacent.

Step 2: Identify the Known and Unknown Quantities

• Write down:
• What is given (e.g., angle = 30°, distance = 50 m).
• What is asked (e.g., "Find the height of the tower").

Step 3: Choose the Right Trigonometric Ratio

• If you have:
• Angle + adjacent side (distance) → height (opposite) → Use tan θ = opposite / adjacent.
• Angle + hypotenuse (line of sight) → height (opposite) → Use sin θ = opposite / hypotenuse.
• Angle + opposite (height) → distance (adjacent) → Use tan θ = opposite / adjacent (rearranged).

Rule of Thumb: If the problem involves height and distance, tan is usually the best choice.

Step 4: Plug Values into the Formula and Solve

• Substitute the known values into the formula.
• Rearrange to solve for the unknown (e.g., height = distance × tan θ).
• Use a calculator (ensure it’s in degree mode for angles in degrees).

Step 5: Check Units and Reasonableness

• Ensure all units match (e.g., meters, not centimeters).
• Ask: "Does this answer make sense?" (e.g., a 50 m tall tree is reasonable; 500 m is not).

Step 6: Write the Final Answer with Units

• Always include units (e.g., "The height of the tower is 28.9 m").

WORKED EXAMPLE (Using the Steps)

Example 1 – Basic (Angle of Elevation)

Problem: From a point 30 m away from the base of a tree, the angle of elevation to the top of the tree is 40°. Find the height of the tree.

Step 1: Draw the Diagram

        Top of tree (T)

           

          /|

         / |

        /  |

       /   |

      /    |

     /     |

    / 40°  |
Observer (O)------- Base of tree (B)

       30 m

• O = Observer
• B = Base of tree
• T = Top of tree
• OB = 30 m (horizontal distance)
• ∠TOB = 40° (angle of elevation)
• BT = ? (height of tree)

Step 2: Identify Known and Unknown

• Given: OB = 30 m, ∠TOB = 40°
• Find: BT (height)

Step 3: Choose the Right Ratio

• We have the adjacent side (OB = 30 m) and need the opposite side (BT = height).
• tan θ = opposite / adjacent → tan 40° = BT / 30

Step 4: Plug in Values and Solve

• BT = 30 × tan 40°
• tan 40° ≈ 0.8391 (use calculator)
• BT = 30 × 0.8391 ≈ 25.17 m

Step 5: Check Reasonableness

• 25 m is a reasonable height for a tree.
• Units are consistent (meters).

Step 6: Final Answer

The height of the tree is 25.2 m (to 1 decimal place).

What we did and why: - We used tan because we had the adjacent side (distance) and needed the opposite side (height). - Drawing the diagram helped us see which sides were which.

Example 2 – Medium (Angle of Depression)

Problem: From the top of a 50 m tall cliff, the angle of depression to a boat at sea is 25°. How far is the boat from the base of the cliff?

Step 1: Draw the Diagram

Cliff top (C)

   

    \

     \ 25° (angle of depression)

      \

       \

         Boat (B)

       /|

      / |

     /  |

    /   |

   /____|
Base (A)

• C = Cliff top
• A = Base of cliff
• B = Boat
• CA = 50 m (height of cliff)
• ∠CBA = 25° (angle of depression = angle of elevation from boat to cliff top)
• AB = ? (distance from boat to base of cliff)

Key Insight: The angle of depression from C to B is equal to the angle of elevation from B to C (alternate angles).

Step 2: Identify Known and Unknown

• Given: CA = 50 m, ∠CBA = 25°
• Find: AB (distance)

Step 3: Choose the Right Ratio

• We have the opposite side (CA = 50 m) and need the adjacent side (AB = distance).
• tan θ = opposite / adjacent → tan 25° = 50 / AB

Step 4: Plug in Values and Solve

• AB = 50 / tan 25°
• tan 25° ≈ 0.4663
• AB = 50 / 0.4663 ≈ 107.2 m

Step 5: Check Reasonableness

• 107 m is a reasonable distance for a boat from a cliff.
• Units are consistent.

Step 6: Final Answer

The boat is 107.2 m away from the base of the cliff.

What we did and why: - We recognized that the angle of depression equals the angle of elevation from the boat. - Used tan because we had the opposite side (height) and needed the adjacent side (distance).

Example 3 – Exam Style (Disguised Problem)

Problem: A hot-air balloon is hovering at a height of 120 m. From a point on the ground, the angle of elevation to the balloon is 60°. How far is the point on the ground from the point directly below the balloon?

Step 1: Draw the Diagram

Balloon (B)

   

    \

     \

      \ 60°

       \

        \

          Point on ground (P)

        /|

       / |

      /  |

     /   |

    /____|
Directly below (D)

• B = Balloon
• D = Point directly below balloon
• P = Observer on ground
• BD = 120 m (height)
• ∠BPD = 60° (angle of elevation)
• PD = ? (distance from observer to point below balloon)

Step 2: Identify Known and Unknown

• Given: BD = 120 m, ∠BPD = 60°
• Find: PD (distance)

Step 3: Choose the Right Ratio

• We have the opposite side (BD = 120 m) and need the adjacent side (PD = distance).
• tan θ = opposite / adjacent → tan 60° = 120 / PD

Step 4: Plug in Values and Solve

• PD = 120 / tan 60°
• tan 60° = √3 ≈ 1.732
• PD = 120 / 1.732 ≈ 69.3 m

Step 5: Check Reasonableness

• 69 m is a reasonable distance.
• Units are consistent.

Step 6: Final Answer

The point on the ground is 69.3 m away from the point directly below the balloon.

What we did and why: - The problem was disguised as a "hot-air balloon" scenario, but it’s just a right-angled triangle. - We used tan because we had the opposite side (height) and needed the adjacent side (distance).

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in—here’s what you must remember for height and distance problems:

1. Draw the diagram first. Label the observer, the object, the angle, and the sides. If it’s an angle of depression, remember it’s the same as the angle of elevation from the other side.
2. Pick the right ratio. If you have height and distance, tan is your best friend. If you have the hypotenuse, use sin or cos.
3. Plug in the numbers carefully. Make sure your calculator is in degree mode, and don’t round until the very end.
4. Check your answer. Does it make sense? A 10 m tall building is reasonable; 1000 m is not.
5. Watch out for traps. Hidden right angles, radians, and multi-step problems are common—don’t rush!

You’ve got this. Practice a few problems tonight, and you’ll be ready to crush this on exam day. Good luck!