How to Solve: Angle of Depression

How to Solve: Angle of Depression

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"You’re standing on a cliff, looking down at a boat in the water. The exam asks: ‘How far is the boat from the base of the cliff?’ Mastering the angle of depression lets you solve this in under 60 seconds—no guesswork, just full marks."

What You Need To Know First

Before diving in, ensure you understand: 1. Right-angled triangles – Sides (opposite, adjacent, hypotenuse) and the Pythagorean theorem. 2. Trigonometric ratios – Sine, cosine, and tangent (SOH-CAH-TOA). 3. Angle of elevation – The angle upward from the horizontal line of sight.

If any of these are shaky, pause here and review them first.

Key Vocabulary

Formulas To Know

1. Tangent Ratio (Most Common for Angle of Depression)

Formula: [ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} ]

Variables: - (\theta) = angle of depression (or the alternate angle inside the triangle). - Opposite = vertical distance (height) from the observer to the object. - Adjacent = horizontal distance from the observer to the object.

Memorise? ✅ MEMORISE THIS (SOH-CAH-TOA is always needed).

2. Sine or Cosine Ratios (If Hypotenuse is Given)

Formulas: [ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} ] [ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} ]

When to use: - If the problem gives the hypotenuse (e.g., a rope or cable length) instead of the horizontal/vertical distance.

Memorise? ✅ MEMORISE THIS (but tangent is more common for angle of depression).

3. Alternate Angles Property

Key Rule: The angle of depression (above the horizontal) is equal to the angle of elevation (below the horizontal) because they are alternate angles.

Why it matters: - You’ll always redraw the problem as a right-angled triangle where the angle of depression becomes an angle of elevation inside the triangle.

Memorise? ✅ MEMORISE THIS (critical for setting up the problem correctly).

Step-by-Step Method

Follow these steps exactly for every angle of depression problem.

Step 1: Draw the Scenario

• Sketch a horizontal line (your line of sight when looking straight ahead).
• Draw the observer (e.g., you on a cliff) and the object (e.g., a boat) below.
• Label the angle of depression (the angle down from the horizontal to the object).

Step 2: Identify the Right Triangle

• From the observer, draw a vertical line down to the level of the object.
• From the object, draw a horizontal line back to the base of the vertical line.
• This forms a right-angled triangle with:
• The angle of depression outside the triangle.
• The angle of elevation inside the triangle (equal to the angle of depression).

Step 3: Label the Triangle

• Opposite side = vertical distance (height) from observer to object.
• Adjacent side = horizontal distance from observer to object.
• Hypotenuse = line of sight (only needed if using sine/cosine).

Step 4: Choose the Right Trig Ratio

• If you have opposite and adjacent, use tangent.
• If you have opposite and hypotenuse, use sine.
• If you have adjacent and hypotenuse, use cosine.

Step 5: Solve for the Unknown

• Plug the known values into the formula.
• Rearrange to solve for the unknown (e.g., distance, height, or angle).
• Use inverse trig functions (e.g., (\tan^{-1})) if solving for an angle.

Step 6: Check Units and Reasonableness

• Ensure all units match (e.g., meters, feet).
• Ask: Does this answer make sense? (e.g., a boat 500m from a cliff is reasonable; 5m is not).

WORKED EXAMPLE (Using the Steps)

Example 1 – Basic

Problem: A person stands on a cliff 50 meters high and looks down at a boat. The angle of depression to the boat is 30°. How far is the boat from the base of the cliff?

Step 1: Draw the Scenario

• Horizontal line from the person’s eyes.
• Angle of depression = 30° (down to the boat).
• Vertical height = 50m.

Step 2: Identify the Right Triangle

• Draw a vertical line from the person to the water level (50m).
• Draw a horizontal line from the boat to the base of the cliff.
• The angle inside the triangle = 30° (alternate angles).

Step 3: Label the Triangle

• Opposite = 50m (height).
• Adjacent = ? (distance from cliff to boat, let’s call this (x)).
• Angle = 30°.

Step 4: Choose the Trig Ratio

• We have opposite (50m) and need adjacent ((x)).
• Use tangent: [ \tan(30°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{50}{x} ]

Step 5: Solve for (x)

[ \tan(30°) = \frac{50}{x} ] [ x = \frac{50}{\tan(30°)} ] [ \tan(30°) = \frac{1}{\sqrt{3}} \approx 0.577 ] [ x = \frac{50}{0.577} \approx 86.6 \text{ meters} ]

Step 6: Check

• 86.6m is reasonable for a boat’s distance from a 50m cliff.
• Units are consistent (meters).

Answer: The boat is 86.6 meters from the base of the cliff.

What we did and why: - We used the angle of depression to find the alternate angle inside the triangle. - Tangent was the best ratio because we had the opposite side and needed the adjacent side.

Example 2 – Medium (Missing Height)

Problem: From the top of a lighthouse, the angle of depression to a ship is 25°. The ship is 120 meters from the base of the lighthouse. How tall is the lighthouse?

Step 1: Draw the Scenario

• Horizontal line from the top of the lighthouse.
• Angle of depression = 25° (down to the ship).
• Horizontal distance = 120m.

Step 2: Identify the Right Triangle

• Draw a vertical line from the lighthouse top to the water (height = (h)).
• Draw a horizontal line from the ship to the lighthouse base (120m).
• Angle inside the triangle = 25° (alternate angles).

Step 3: Label the Triangle

• Opposite = (h) (height of lighthouse).
• Adjacent = 120m.
• Angle = 25°.

Step 4: Choose the Trig Ratio

• We have adjacent (120m) and need opposite ((h)).
• Use tangent: [ \tan(25°) = \frac{h}{120} ]

Step 5: Solve for (h)

[ h = 120 \times \tan(25°) ] [ \tan(25°) \approx 0.466 ] [ h = 120 \times 0.466 \approx 55.9 \text{ meters} ]

Step 6: Check

• 55.9m is a reasonable height for a lighthouse.
• Units are consistent (meters).

Answer: The lighthouse is 55.9 meters tall.

What we did and why: - We used the given horizontal distance and angle to find the height. - Tangent was the best ratio because we had the adjacent side and needed the opposite side.

Example 3 – Exam Style (Disguised Problem)

Problem: A drone hovers 80 meters above the ground. The angle of depression to a lost hiker is 40°. How far is the hiker from the point on the ground directly below the drone? Give your answer to 1 decimal place.

Step 1: Draw the Scenario

• Horizontal line from the drone.
• Angle of depression = 40° (down to the hiker).
• Vertical height = 80m.

Step 2: Identify the Right Triangle

• Draw a vertical line from the drone to the ground (80m).
• Draw a horizontal line from the hiker to the point below the drone ((x)).
• Angle inside the triangle = 40° (alternate angles).

Step 3: Label the Triangle

• Opposite = 80m.
• Adjacent = (x).
• Angle = 40°.

Step 4: Choose the Trig Ratio

• We have opposite (80m) and need adjacent ((x)).
• Use tangent: [ \tan(40°) = \frac{80}{x} ]

Step 5: Solve for (x)

[ x = \frac{80}{\tan(40°)} ] [ \tan(40°) \approx 0.839 ] [ x = \frac{80}{0.839} \approx 95.4 \text{ meters} ]

Step 6: Check

• 95.4m is a reasonable distance for a hiker from a drone.
• Answer is to 1 decimal place as requested.

Answer: The hiker is 95.4 meters from the point directly below the drone.

What we did and why: - The problem was disguised as a drone scenario, but it’s the same as a cliff/boat problem. - We used tangent because we had the opposite side and needed the adjacent side.

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in. The angle of depression is the angle down from your eye level to an object. But here’s the trick: it’s equal to the angle of elevation inside the triangle you draw. So, sketch the scenario, label the right triangle, and use SOH-CAH-TOA—usually tangent—to solve for the missing side. Double-check your opposite and adjacent sides, and always ask: ‘Does this answer make sense?’ If you’re on a 50m cliff, a boat 10m away is impossible, but 80m is realistic. Practice two problems tonight, and you’ll own this on exam day. You’ve got this!