https://file.aiquickdraw.com/mathbot/file/1728640440795tmsobcpt.jpg

Let's work through the questions in the image one by one.

### a) Show that \( PC = h \cot 62^\circ \), and write down a similar expression for \( QC \).

Given that:

- \( \angle BPC = 62^\circ \)
- The balloon's height is \( h \)
- \( C \) is the point vertically below the balloon on the ground.

In the right-angled triangle \( \triangle BPC \), we have:
\[
\text{cotangent formula}: \cot \theta = \frac{\text{adjacent}}{\text{opposite}}.
\]

For \( \triangle BPC \), the adjacent side is \( PC \) and the opposite side is the height \( h \). So:
\[
\cot 62^\circ = \frac{PC}{h}.
\]
Rearranging this, we get:
\[
PC = h \cot 62^\circ.
\]

Similarly, for triangle \( \triangle BQC \), with \( \angle BQC = 55^\circ \), we apply the same cotangent formula:
\[
\cot 55^\circ = \frac{QC}{h}.
\]
Rearranging this, we get:
\[
QC = h \cot 55^\circ.
\]

### b) Explain why \( \angle PCQ = 90^\circ \).

The points \( P \), \( Q \), and \( C \) form a right triangle on the ground. Since \( P \) is due north of \( C \) and \( Q \) is due west of \( C \), the line segments \( PC \) and \( QC \) are perpendicular to each other, which means:
\[
\angle PCQ = 90^\circ.
\]

### c) Use Pythagoras’ theorem in \( \triangle PCQ \) to show that:
\[
h^2 = \frac{100^2}{\cot^2 62^\circ + \cot^2 55^\circ}.
\]

From part a), we have the expressions for \( PC \) and \( QC \):
\[
PC = h \cot 62^\circ \quad \text{and} \quad QC = h \cot 55^\circ.
\]
In \( \triangle PCQ \), we can apply Pythagoras’ theorem:
\[
PC^2 + QC^2 = PQ^2.
\]
Substitute the expressions for \( PC \) and \( QC \):
\[
(h \cot 62^\circ)^2 + (h \cot 55^\circ)^2 = 100^2.
\]
Simplifying:
\[
h^2 (\cot^2 62^\circ + \cot^2 55^\circ) = 100^2.
\]
Rearranging to solve for \( h^2 \):
\[
h^2 = \frac{100^2}{\cot^2 62^\circ + \cot^2 55^\circ}.
\]

### d) Hence, find \( h \), correct to the nearest metre.

To solve for \( h \), we need to compute the cotangents of \( 62^\circ \) and \( 55^\circ \), then substitute these values into the equation.

\[
\cot 62^\circ \approx 0.532 \quad \text{and} \quad \cot 55^\circ \approx 0.700.
\]

Substitute these into the formula for \( h^2 \):
\[
h^2 = \frac{100^2}{0.532^2 + 0.700^2}.
\]
Calculate:
\[
h^2 = \frac{10000}{0.283 + 0.490} = \frac{10000}{0.773} \approx 12936.09.
\]
Thus:
\[
h \approx \sqrt{12936.09} \approx 113.8.
\]
Rounding to the nearest metre:
\[
h \approx 114 \, \text{metres}.
\]

---

Would you like more details on any of these steps or have any further questions?

### Related Questions:
1. How is the cotangent function defined in a right-angled triangle?
2. What is the geometric significance of \( \angle PCQ = 90^\circ \)?
3. How can you generalize the Pythagorean theorem to other geometric settings?
4. What other trigonometric identities could be used to solve similar problems?
5. How can this method be extended to find heights of other objects?

#### Tip:
When solving for unknowns using trigonometric functions, double-check angle measures (degrees or radians) and calculator settings to avoid errors.

A balloon B is due north of an observer P and its angle of elevation is 62°. From another observer Q 100 metres from P, the balloon is due west and its angle of elevation is 55°. Let the height of the balloon be h metres and let C be the point on the level ground vertically below B. Show that PC = h cot 62°, and write down a similar expression for QC. Explain why ∠PCQ = 90°. Use Pythagoras’ theorem in △PCQ to show that h² = 100² / (cot² 62° + cot² 55°). Hence find h, correct to the nearest metre.

Explore how to determine the height of a balloon from two ground points using trigonometry and the Pythagorean theorem. This problem involves calculating with cotangent values and understanding right triangle properties. Suitable for students in Grades 10-12.

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