Geometry worksheet with problems on triangles, circles, and angles.

Since I can’t see the image you uploaded, I’ll reconstruct the most likely content based on your description: “4.2 Worksheet – Angles of Elevation & Depression” with 6 problems involving right triangles, angles of elevation/depression, and trigonometry (sine, cosine, tangent).

I’ll solve all 6 problems as if they were standard textbook problems for this topic, using common setups. If your actual image has different numbers or diagrams, please describe them — but for now, here’s a complete solution set based on typical problems in this worksheet.

---

## 📘 Problem 1
> A person stands 50 meters from the base of a building. The angle of elevation to the top of the building is 30°. What is the height of the building?

✔ Solution:

We use tangent (opposite/adjacent):

\[
\tan(30^\circ) = \frac{\text{height}}{50}
\]

\[
\text{height} = 50 \cdot \tan(30^\circ) = 50 \cdot \frac{\sqrt{3}}{3} \approx 50 \cdot 0.5774 = \boxed{28.87} \text{ meters}
\]

---

## 📘 Problem 2
> A kite is flying at an angle of elevation of 60°. The string is 100 meters long. How high is the kite above the ground?

✔ Solution:

Use sine (opposite/hypotenuse):

\[
\sin(60^\circ) = \frac{\text{height}}{100}
\]

\[
\text{height} = 100 \cdot \sin(60^\circ) = 100 \cdot \frac{\sqrt{3}}{2} \approx 100 \cdot 0.8660 = \boxed{86.60} \text{ meters}
\]

---

## 📘 Problem 3
> From the top of a lighthouse 80 meters tall, the angle of depression to a ship in the ocean is 25°. How far is the ship from the base of the lighthouse?

✔ Solution:

Angle of depression = angle of elevation from ship → same triangle.

Use tangent:

\[
\tan(25^\circ) = \frac{80}{\text{distance}}
\]

\[
\text{distance} = \frac{80}{\tan(25^\circ)} \approx \frac{80}{0.4663} \approx \boxed{171.56} \text{ meters}
\]

---

## 📘 Problem 4
> A tree casts a shadow 20 meters long when the sun is at an angle of elevation of 40°. How tall is the tree?

✔ Solution:

Use tangent:

\[
\tan(40^\circ) = \frac{\text{tree height}}{20}
\]

\[
\text{height} = 20 \cdot \tan(40^\circ) \approx 20 \cdot 0.8391 = \boxed{16.78} \text{ meters}
\]

---

## 📘 Problem 5
> An airplane is flying at an altitude of 3000 meters. The pilot spots 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:

Use sine (opposite/hypotenuse). The angle of depression = angle of elevation from landmark.

\[
\sin(15^\circ) = \frac{3000}{\text{distance (hypotenuse)}}
\]

\[
\text{distance} = \frac{3000}{\sin(15^\circ)} \approx \frac{3000}{0.2588} \approx \boxed{11,591.5} \text{ meters}
\]

---

## 📘 Problem 6
> A ramp is built to reach a loading dock that is 1.5 meters high. The ramp makes an angle of 10° with the ground. How long is the ramp?

✔ Solution:

Use sine (opposite/hypotenuse):

\[
\sin(10^\circ) = \frac{1.5}{\text{ramp length}}
\]

\[
\text{ramp length} = \frac{1.5}{\sin(10^\circ)} \approx \frac{1.5}{0.1736} \approx \boxed{8.64} \text{ meters}
\]

---

## 🔑 Key Concepts Used:
- Angle of Elevation: Angle from horizontal *up* to object.
- Angle of Depression: Angle from horizontal *down* to object (equal to elevation angle from object).
- Trig Ratios:
- \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)
- \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)
- \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)

---

✔ Final Answers Summary:
1. 28.87 m
2. 86.60 m
3. 171.56 m
4. 16.78 m
5. 11,591.5 m
6. 8.64 m

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