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Let's break down the problems shown in the image and solve them step by step.

### Problem 18:

**Given:**
- A tree is 2 meters high.
- A tower is 8 meters high.
- Point \( P \) is 3 meters away from the tree.
- The angles of elevation to the top of the tree and the top of the tower are equal.
- We are to find the distance \( x \) from \( P \) to the top of the tower.

**Solution:**
Since the angles of elevation to the top of the tree and the top of the tower are equal, we can use the concept of similar triangles.

Let the distance from \( P \) to the base of the tower be \( d \). The heights of the tree and the tower are given, so we can set up the following ratio based on similar triangles:

\[
\frac{\text{Height of the Tree}}{\text{Distance to the Tree}} = \frac{\text{Height of the Tower}}{\text{Distance to the Tower}}
\]

Substituting the known values:

\[
\frac{2}{3} = \frac{8}{d}
\]

Now, solve for \( d \):

\[
d = \frac{8 \times 3}{2} = 12 \text{ meters}
\]

The total distance from \( P \) to the top of the tower, \( x \), is calculated using the Pythagorean theorem:

\[
x = \sqrt{d^2 + (\text{Height of the Tower})^2} = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.42 \text{ meters}
\]

So the answer is **(D) 14.42 m**.

---

### Problem 19:

**Given:**
- A sphere and a closed cylinder have the same radius.
- The height of the cylinder is four times the radius.
- We are asked to find the ratio of the volume of the cylinder to the volume of the sphere.

**Solution:**

1. **Volume of a sphere:**

   \[
   V_{\text{sphere}} = \frac{4}{3} \pi r^3
   \]

2. **Volume of a cylinder:**

   \[
   V_{\text{cylinder}} = \pi r^2 h
   \]

   Given that the height \( h = 4r \):

   \[
   V_{\text{cylinder}} = \pi r^2 \times 4r = 4\pi r^3
   \]

3. **Ratio of the volumes:**

   \[
   \text{Ratio} = \frac{V_{\text{cylinder}}}{V_{\text{sphere}}} = \frac{4\pi r^3}{\frac{4}{3} \pi r^3} = \frac{4}{\frac{4}{3}} = 3
   \]

So, the ratio is **(C) 3**.

---

### Problem 20:

**Given:**
- \( Q \) is SW of \( P \).
- \( R \) is NW of \( P \).
- \( \angle QPT = 165^\circ \).
- We are asked to find the bearing of \( T \) from \( P \).

**Solution:**

The bearing is measured clockwise from the north.

1. \( Q \) is southwest (SW) of \( P \), so its bearing is 225°.
2. \( R \) is northwest (NW) of \( P \), so its bearing is 315°.
3. The angle \( \angle QPT = 165^\circ \).

To calculate the bearing of \( T \) from \( P \), we can subtract the angle \( \angle QPT \) from the bearing of \( Q \):

\[
\text{Bearing of } T = 225^\circ - 165^\circ = 60^\circ
\]

So the answer is **(A) 060°**.

---

### Summary of Answers:
- **18:** (D) 14.42 m
- **19:** (C) 3
- **20:** (A) 060°

Would you like further details or clarification on any of these solutions? 

Here are 5 related questions you can explore:
1. How would the answer to Problem 18 change if the tower were 10 meters high instead of 8 meters?
2. What would be the volume ratio if the cylinder's height were twice the radius instead of four times in Problem 19?
3. How would the bearing change in Problem 20 if \( \angle QPT \) were 120°?
4. How do you determine the area of a triangle if the angles of elevation from two points are known?
5. How do similar triangles apply to problems involving heights and distances?

**Tip:** When solving geometry problems involving angles of elevation, try drawing a diagram to visualize the relationships between the distances and angles clearly.

solve the problems in the provided math exam sheet

Explore detailed solutions for a math exam sheet involving problems with similar triangles, Pythagorean theorem, volume calculations of spheres and cylinders, and bearing calculations. This guide is ideal for high school students looking to understand complex geometry concepts and calculations.

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