Word Problems Angles of Elevation and Depression | Lecture notes ... - Free Printable

Let's go through each of the word problems involving angles of elevation and depression, verify the solutions, and explain the reasoning step-by-step.

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Problem 1:

> A woman is standing on the ground at a point 78 ft from the base of a building. The angle of elevation to the top of the building is 57°. To the nearest foot, how high is the building?

#### Given:
- Distance from observer to building (adjacent side) = 78 ft
- Angle of elevation = 57°
- Height of building = $ h $ (opposite side)

We use the tangent function:
$$
\tan(57^\circ) = \frac{h}{78}
$$

Solve for $ h $:
$$
h = 78 \cdot \tan(57^\circ)
$$

Using a calculator:
$$
\tan(57^\circ) \approx 1.5399
$$
$$
h \approx 78 \cdot 1.5399 \approx 120.01 \text{ ft}
$$

✔ Answer: $ \boxed{120} $ ft (to the nearest foot)

✔️ Correct.

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Problem 2:

> The sun shines on a flagpole, causing a shadow to be cast on the ground. The distance from the base of the pole to the tip of the shadow is 49 feet. At that time of day, the sun’s rays make an angle of 38° with the ground. How tall is the flagpole?

#### Given:
- Shadow length (adjacent) = 49 ft
- Angle of elevation of sun = 38°
- Height of flagpole = $ h $

Use tangent:
$$
\tan(38^\circ) = \frac{h}{49}
\Rightarrow h = 49 \cdot \tan(38^\circ)
$$

$$
\tan(38^\circ) \approx 0.7813
\Rightarrow h \approx 49 \cdot 0.7813 \approx 38.28 \text{ ft}
$$

✔ Answer: $ \boxed{38.3} $ ft (rounded to nearest tenth)

✔️ Correct.

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Problem 3:

> An observer on a cliff 1000 dm above sea level sights two ships due east. The angles of depression of the ships are 47° and 32°. Find, to the nearest decimeter, the distance between the two ships.

#### Understanding:
- Observer is at height = 1000 dm
- Angles of depression to two ships are 47° and 32°
- Since the angles are measured from the horizontal downward, they form right triangles with the vertical cliff.

Let:
- $ x $ = horizontal distance from observer to closer ship (angle 47°)
- $ y $ = horizontal distance to farther ship (angle 32°)

Then:
$$
\tan(47^\circ) = \frac{1000}{x} \Rightarrow x = \frac{1000}{\tan(47^\circ)}
$$
$$
\tan(32^\circ) = \frac{1000}{y} \Rightarrow y = \frac{1000}{\tan(32^\circ)}
$$

Now calculate:

- $ \tan(47^\circ) \approx 1.0724 $
$$
x \approx \frac{1000}{1.0724} \approx 932.3 \text{ dm}
$$

- $ \tan(32^\circ) \approx 0.6249 $
$$
y \approx \frac{1000}{0.6249} \approx 1599.6 \text{ dm}
$$

Distance between ships = $ y - x \approx 1599.6 - 932.3 = 667.3 $ dm

Rounded to nearest decimeter: $ \boxed{667} $ dm

But in the image, it says 668 dm — very close.

Let’s check precision:
- More accurate values:
- $ \tan(47^\circ) \approx 1.072365 $
$ x \approx 1000 / 1.072365 \approx 932.32 $
- $ \tan(32^\circ) \approx 0.624869 $
$ y \approx 1000 / 0.624869 \approx 1599.63 $
- Difference: $ 1599.63 - 932.32 = 667.31 $ → rounds to 667 dm

✘ So 668 dm is slightly off. It should be 667 dm.

However, if using approximated values:
- If $ \tan(47^\circ) \approx 1.07 $ → $ x \approx 934.6 $
- $ \tan(32^\circ) \approx 0.62 $ → $ y \approx 1612.9 $
- Diff ≈ 678.3 → not matching

Wait — perhaps the student used more precise intermediate steps.

Let’s double-check the student's work:
- They wrote:
$$
y = \frac{1000}{\tan(32^\circ)} \quad x = \frac{1000}{\tan(47^\circ)}
$$
Then:
$$
\text{Distance} = y - x = \frac{1000}{\tan(32^\circ)} - \frac{1000}{\tan(47^\circ)}
$$

Plug in:
- $ \tan(32^\circ) \approx 0.6249 $ → $ 1000 / 0.6249 \approx 1599.6 $
- $ \tan(47^\circ) \approx 1.0724 $ → $ 1000 / 1.0724 \approx 932.3 $
- $ 1599.6 - 932.3 = 667.3 $

So correct answer is 667 dm.

But student wrote 668 dm — likely a rounding error or calculator difference.

✔ Correct Answer: $ \boxed{667} $ dm
✘ Student’s answer is slightly off.

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Problem 4:

> A 200 ft high television transmitting tower is to be supported by guy wires running from the ground to the top of the tower. The wires make an angle of 63° with the ground.
>
> a) How long is each wire?
> b) How far from the base of the tower must they meet the ground?

#### Diagram:
Right triangle:
- Opposite side = 200 ft (height)
- Angle = 63°
- Hypotenuse = $ w $ (wire length)
- Adjacent = $ b $ (base distance)

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#### Part a): Length of wire (hypotenuse)

Use sine:
$$
\sin(63^\circ) = \frac{200}{w}
\Rightarrow w = \frac{200}{\sin(63^\circ)}
$$

$ \sin(63^\circ) \approx 0.8910 $

$$
w \approx \frac{200}{0.8910} \approx 224.46 \text{ ft}
\Rightarrow \boxed{224.5} \text{ ft}
$$

✔ Correct.

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#### Part b): Distance from base (adjacent side)

Use tangent:
$$
\tan(63^\circ) = \frac{200}{b}
\Rightarrow b = \frac{200}{\tan(63^\circ)}
$$

$ \tan(63^\circ) \approx 1.9626 $

$$
b \approx \frac{200}{1.9626} \approx 101.91 \text{ ft}
\Rightarrow \boxed{101.9} \text{ ft}
$$

✔ Correct.

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \boxed{120} $ ft |
| 2 | $ \boxed{38.3} $ ft |
| 3 | $ \boxed{667} $ dm *(not 668)* |
| 4a | $ \boxed{224.5} $ ft |
| 4b | $ \boxed{101.9} $ ft |

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🔍 Explanation of Key Concepts:

- Angle of Elevation: When looking up from horizontal to an object (e.g., top of building).
- Angle of Depression: When looking down from horizontal to an object (e.g., ship from cliff).
- Use SOHCAHTOA:
- $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $
- $ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} $
- $ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} $
- Always draw a sketch!
- Be careful with units (dm vs ft).

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🟡 Note:

In Problem 3, the student's answer of 668 dm is close but incorrect due to rounding. The exact calculation gives 667 dm. This may be due to calculator precision or rounding during steps.

✔ All other answers are correct and well-explained.

Let me know if you'd like a diagram or further clarification!