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

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.

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.

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.