https://file.aiquickdraw.com/mathbot/file/1731829742973qmow9k9e.jpg

Let's solve the two problems shown in the uploaded image step-by-step.

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### **Problem 10: Finding the Height of the Tree**
We are given:
1. Point \( A \), where the angle of elevation to the tree's top is \( 51^\circ \).
2. Point \( B \), 25 ft away from \( A \), where the angle \( \angle ABC = 40^\circ \).

To find the height \( h \), we can use trigonometry and the Law of Sines.

#### Steps:
1. Label the triangle:
   - \( \triangle ABC \), with \( AB = 25 \, \text{ft} \).
   - \( \angle A = 51^\circ \) and \( \angle B = 40^\circ \).
   - \( \angle C = 180^\circ - 51^\circ - 40^\circ = 89^\circ \).

2. Apply the Law of Sines to find the hypotenuse \( AC \):
   \[
   \frac{AC}{\sin 40^\circ} = \frac{25}{\sin 89^\circ}
   \]
   Since \( \sin 89^\circ \approx 1 \):
   \[
   AC = 25 \times \sin 40^\circ \approx 25 \times 0.6428 = 16.07 \, \text{ft}.
   \]

3. Use trigonometry to find \( h \) using \( \angle A \) (angle of elevation):
   \[
   h = AC \times \sin 51^\circ
   \]
   \[
   h \approx 16.07 \times 0.7771 = 12.49 \, \text{ft}.
   \]

**Answer for Problem 10:**
The height of the tree is approximately **12 feet**.

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### **Problem 11: Distance of the Ship from Two Points**
We are given:
1. Two observation points 18 miles apart along a shoreline.
2. Bearings:
   - From the northern point: \( N32^\circ E \),
   - From the southern point: \( S58^\circ E \).

This forms a triangle with:
1. The base \( AB = 18 \, \text{mi} \) between the observation points.
2. Interior angles:
   - \( \angle A = 180^\circ - 32^\circ = 148^\circ \) (angle at the northern point),
   - \( \angle B = 180^\circ - 58^\circ = 122^\circ \),
   - \( \angle C = 180^\circ - 148^\circ - 122^\circ = 10^\circ \).

#### Steps:
1. Apply the Law of Sines to find the distances \( AC \) and \( BC \):
   \[
   \frac{AC}{\sin 122^\circ} = \frac{BC}{\sin 148^\circ} = \frac{18}{\sin 10^\circ}.
   \]

2. Solve for \( AC \) and \( BC \):
   - \( AC = \frac{18 \times \sin 122^\circ}{\sin 10^\circ} \),
   - \( BC = \frac{18 \times \sin 148^\circ}{\sin 10^\circ} \).

3. Approximate values:
   - \( \sin 122^\circ \approx 0.848 \),
   - \( \sin 148^\circ \approx 0.629 \),
   - \( \sin 10^\circ \approx 0.174 \).

   Compute:
   - \( AC \approx \frac{18 \times 0.848}{0.174} \approx 87.69 \, \text{mi} \),
   - \( BC \approx \frac{18 \times 0.629}{0.174} \approx 65.05 \, \text{mi} \).

**Answer for Problem 11:**
- Distance from northern point: **87.7 miles**.
- Distance from southern point: **65.1 miles**.

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Do you want detailed breakdowns of the calculations or further clarifications?  

**Follow-Up Questions:**
1. Would you like to see the Law of Sines derivation for both problems?
2. Should I calculate using higher precision for Problem 11?
3. Do you want a visual representation of the triangles in both problems?
4. Would you like me to explain how bearings are used in trigonometric navigation problems?
5. Should I round the answers differently?

**Tip:** Always label angles and sides clearly in geometry problems to avoid confusion when applying trigonometric formulas.

Two problems from the provided image:
1. Estimating the height of a tree using trigonometry based on angles of elevation and distance measurements.
2. Determining the distances of a ship from two observation points based on bearings and the Law of Sines.

Solve two real-world trigonometry problems: estimating the height of a tree across a stream using angles of elevation, and calculating the distance of a ship from two points using bearings. Includes detailed use of the Law of Sines and angle calculations. Suitable for Grades 10-12.

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