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### Problem Analysis:
1. A lighthouse (A) has a lookout at the top. The angle of depression from A to a boat at point C is 30°.
2. After 1 minute, the boat travels 50 m towards the lighthouse and is now at point B.
3. The angle of elevation from point B to the lighthouse lookout is 60°.

### Given Data:
- Distance \( BC = 50 \) m.
- Angle of depression from A to C: \( \angle AOC = 30^\circ \).
- Angle of elevation from B to A: \( \angle ABO = 60^\circ \).

### Questions:
(a) Write down the value of \( x^\circ \) in the diagram.

(b) Find the height of the lighthouse.

(c) Find the speed of the boat, in meters per second, from C to B.

### Solution:

#### (a) Value of \( x^\circ \):
The value of \( x^\circ \) is the angle at C, which is 60°. This is because \( \angle AOC = 30^\circ \) and the angle of elevation at B to A is 60°.

**Answer (a):** \( x = 60^\circ \).

#### (b) Height of the lighthouse:
Let the height of the lighthouse be \( h \). From the right triangle \( AOB \):

Using the tangent of the angle at B:
\[
\tan(60^\circ) = \frac{h}{OB}.
\]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[
OB = \frac{h}{\sqrt{3}}.
\]

Now, using the right triangle \( AOC \):

\[
\tan(30^\circ) = \frac{h}{OC},
\]
and since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[
OC = h \cdot \sqrt{3}.
\]

Given that \( OC = OB + BC \):
\[
h \cdot \sqrt{3} = \frac{h}{\sqrt{3}} + 50.
\]

Multiplying the entire equation by \( \sqrt{3} \):
\[
3h = h + 50\sqrt{3}.
\]

Simplify:
\[
2h = 50\sqrt{3},
\]
\[
h = 25\sqrt{3} \approx 25 \times 1.732 = 43.3 \text{ m}.
\]

**Answer (b):** The height of the lighthouse is approximately 43.3 m.

#### (c) Speed of the boat:
The boat travels from C to B, a distance of 50 m, in 1 minute (60 seconds).

Speed:
\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{50 \text{ m}}{60 \text{ s}} \approx 0.833 \text{ m/s}.
\]

**Answer (c):** The speed of the boat is approximately 0.833 m/s.

---

Would you like further details or explanations on any part? Here are some related questions:

1. How would the height of the lighthouse change if the angles were different?
2. Can you find the time taken if the boat was moving at a different speed?
3. What other trigonometric ratios could be used to solve similar problems?
4. How would you find the horizontal distance covered by the boat if it traveled in a different direction?
5. How do changes in elevation angles affect distance measurements in trigonometry?

**Tip:** Always check the units of measurement when calculating speeds, distances, or heights to ensure consistency and accuracy.

The angle of depression from a lookout at the top of a lighthouse (A) down to a boat located at point C is 30°. The boat travels towards the lighthouse and after 1 minute has traveled a distance of 50 m and is now located at point B. The angle of elevation from the boat at B up to the lighthouse lookout is 60°. (a) Write down the value of x° in the diagram. (b) Find the height of the lighthouse. (c) Find the speed of the boat, in meters per second, from C to B.

Solve a trigonometry problem involving a lighthouse and a boat. Given angles of elevation and depression, learn how to calculate the height of the lighthouse and the boat's speed. Suitable for Grades 9-10.

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