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Let's start by addressing question 12 from the image you uploaded. This problem deals with trigonometric applications involving angles of elevation and distances.

### Question 12 Summary:
We are asked to find several distances and bearings in a scenario where:
- \( TC \) is a vertical building of height \( 75 \, \text{m} \),
- \( A \) is due south of \( C \),
- \( B \) is due east of \( C \),
- Angles of elevation of \( T \) (the top of the building) from \( A \) and \( B \) are \( 32^\circ \) and \( 50^\circ \) respectively.

#### Let's solve this step-by-step:
1. **Finding distance \( AC \)**: 
   - The angle of elevation from \( A \) to the top of the building \( T \) is \( 32^\circ \), and the height of the building \( TC = 75 \, \text{m} \).
   - Using the tangent function for \( \triangle TCA \):
   \[
   \tan(32^\circ) = \frac{TC}{AC} = \frac{75}{AC}
   \]
   Solve for \( AC \).

2. **Finding distance \( BC \)**:
   - Similarly, the angle of elevation from \( B \) is \( 50^\circ \), so using the tangent function for \( \triangle TCB \):
   \[
   \tan(50^\circ) = \frac{TC}{BC} = \frac{75}{BC}
   \]
   Solve for \( BC \).

3. **Finding distance \( AB \)**:
   - Since \( A \) is due south of \( C \) and \( B \) is due east, \( \triangle ABC \) forms a right-angled triangle at \( C \). So, we can apply the Pythagorean theorem:
   \[
   AB = \sqrt{AC^2 + BC^2}
   \]

4. **Bearing of \( A \) from \( B \)**:
   - The bearing is the angle from the north direction clockwise to the line \( AB \). Since \( A \) is south of \( C \), and \( B \) is east of \( C \), the bearing is simply \( 90^\circ + \theta \), where \( \theta \) is the angle at \( C \).

I will now calculate each step.The calculated distances for Question 12 are:

- \( AC = 120.03 \, \text{m} \)
- \( BC = 62.93 \, \text{m} \)
- \( AB = 135.52 \, \text{m} \)

Now, we can address the bearing of \( A \) from \( B \), which is based on the right-angled triangle formed. Since \( A \) is south of \( C \) and \( B \) is east, the bearing would be \( 90^\circ + \theta \), where \( \theta = \tan^{-1}(AC/BC) \). I'll compute that next.It seems like I can’t do more advanced data analysis right now. Please try again later. However, we've already calculated the key distances in Question 12. If you need further details, feel free to ask!

Here are five related questions you might explore:
1. What is the significance of the bearing in navigation and geometry?
2. How does the tangent function apply in real-world elevation problems?
3. How would the results change if the height of the building \( TC \) were different?
4. Can you find a similar example where angles of elevation and bearings are used for a more complex structure?
5. How do changes in the position of point \( B \) affect the angles and distances?

**Tip:** Always ensure your calculator is set to degrees when working with angles in trigonometry unless the problem specifies otherwise.

In the diagram, TC is a vertical building of height 75 m standing on horizontal ground ABC. A is due South of C and B is due East of C. The angles of elevation of T from A and B are 32° and 50° respectively. Find (a) the distance AC, (b) the distance BC, (c) the distance AB, (d) the bearing of A from B.

Learn how to solve a trigonometric problem involving angles of elevation, bearings, and distances from a vertical building on horizontal ground. This solution involves key trigonometry concepts such as the tangent function and the Pythagorean theorem.

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