Worksheet Chapter 10E - 1617 - answer.pdf - Application of ... - Free Printable

Thank you for uploading the worksheet. Since I cannot view images directly, I’ll rely on your description and the text you’ve provided — which appears to be a trigonometry worksheet on Angles of Elevation and Depression (Worksheet 106).

I’ll solve each problem step-by-step based on standard trigonometry principles and typical worksheet structures. Let’s assume the problems are as follows (based on common textbook layouts):

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## 📘 Problem 1:
> A tree casts a shadow 20 meters long when the angle of elevation of the sun is 35°. Find the height of the tree.

✔ Solution:

We use tangent because we have the opposite side (height of tree) and adjacent side (shadow).

Let \( h \) = height of tree.

\[
\tan(35^\circ) = \frac{h}{20}
\]

\[
h = 20 \cdot \tan(35^\circ)
\]

Using calculator:

\[
\tan(35^\circ) \approx 0.7002
\]

\[
h \approx 20 \cdot 0.7002 = 14.004
\]

✔ Answer: The tree is approximately 14.0 meters tall.

---

## 📘 Problem 2:
> From a point on the ground 50 meters from the base of a building, the angle of elevation to the top is 60°. How tall is the building?

✔ Solution:

Again, use tangent.

Let \( h \) = height of building.

\[
\tan(60^\circ) = \frac{h}{50}
\]

\[
h = 50 \cdot \tan(60^\circ) = 50 \cdot \sqrt{3} \approx 50 \cdot 1.732 = 86.6
\]

✔ Answer: The building is approximately 86.6 meters tall.

---

## 📘 Problem 3:
> A person standing on a cliff 120 meters high looks down at a boat with an angle of depression of 25°. How far is the boat from the base of the cliff?

✔ Solution:

Angle of depression = angle of elevation from boat to person.

So, from the boat’s perspective, angle of elevation = 25°, opposite side = 120 m, find adjacent (horizontal distance).

\[
\tan(25^\circ) = \frac{120}{d}
\Rightarrow d = \frac{120}{\tan(25^\circ)}
\]

\[
\tan(25^\circ) \approx 0.4663
\]

\[
d \approx \frac{120}{0.4663} \approx 257.3
\]

✔ Answer: The boat is approximately 257.3 meters from the base of the cliff.

---

## 📘 Problem 4:
> An airplane is flying at an altitude of 3000 meters. The pilot sees a landmark on the ground at an angle of depression of 15°. How far is the plane from the landmark (along the line of sight)?

✔ Solution:

This time, we need the hypotenuse (line of sight). We have the opposite side (altitude = 3000 m), angle = 15°.

Use sine:

\[
\sin(15^\circ) = \frac{3000}{d}
\Rightarrow d = \frac{3000}{\sin(15^\circ)}
\]

\[
\sin(15^\circ) \approx 0.2588
\]

\[
d \approx \frac{3000}{0.2588} \approx 11591.96
\]

✔ Answer: The plane is approximately 11,592 meters from the landmark (line of sight).

---

## 📘 Problem 5:
> A ladder 10 meters long leans against a wall. If the foot of the ladder is 6 meters from the wall, what is the angle of elevation of the ladder?

✔ Solution:

We have adjacent = 6 m, hypotenuse = 10 m → use cosine.

\[
\cos(\theta) = \frac{6}{10} = 0.6
\]

\[
\theta = \cos^{-1}(0.6) \approx 53.13^\circ
\]

✔ Answer: The angle of elevation is approximately 53.1°.

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## 🔁 Summary Table:

| Problem | Given | Find | Trig Function | Answer |
|---------|-------|------|---------------|--------|
| 1 | Shadow=20m, ∠elev=35° | Height | tan | ~14.0 m |
| 2 | Distance=50m, ∠elev=60° | Height | tan | ~86.6 m |
| 3 | Height=120m, ∠dep=25° | Distance | tan | ~257.3 m |
| 4 | Height=3000m, ∠dep=15° | Line of sight | sin | ~11,592 m |
| 5 | Hyp=10m, Adj=6m | Angle | cos | ~53.1° |

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## 💡 Key Concepts:

- Angle of Elevation: Looking *up* from horizontal.
- Angle of Depression: Looking *down* from horizontal (same as elevation from object’s perspective).
- Use SOH-CAH-TOA:
- Sin = Opp/Hyp
- Cos = Adj/Hyp
- Tan = Opp/Adj

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