We are tasked with solving a triangle problem using the Law of Sines. The triangle is defined by:

• Points $A$, $B$, and $C$, where $A$ and $B$ are on the shoreline, and $C$ is the sailboat.
• The distance between $A$ and $B$ is $20$ meters.
• Angle $\angle A = 65^\circ$.
• Angle $\angle B = 42^\circ$.

Step 1: Find $\angle C$

Since the sum of the angles in any triangle is $180^\circ$:

$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 65^\circ - 42^\circ = 73^\circ$

Step 2: Apply the Law of Sines

We can use the Law of Sines to find the distances from point $C$ to $A$ and $B$, denoted as $AC$ and $BC$.

The Law of Sines states:

$\frac{AC}{\sin(\angle B)} = \frac{BC}{\sin(\angle A)} = \frac{AB}{\sin(\angle C)}$

Let's first find the distance $AC$ and $BC$ using the known side $AB = 20$ meters.

Solving for $AC$:

$\frac{AC}{\sin(42^\circ)} = \frac{20}{\sin(73^\circ)}$

Solving this for $AC$:

$AC = \frac{20 \cdot \sin(42^\circ)}{\sin(73^\circ)}$

Solving for $BC$:

$\frac{BC}{\sin(65^\circ)} = \frac{20}{\sin(73^\circ)}$

Solving this for $BC$:

$BC = \frac{20 \cdot \sin(65^\circ)}{\sin(73^\circ)}$

I will now calculate these values.The distances from the sailboat at point $C$ to the two observers are:

• Distance from $C$ to $A$: $13.99 \, \text{meters}$
• Distance from $C$ to $B$: $18.95 \, \text{meters}$

Would you like more details on how the calculations were performed or clarification on any steps?

Here are 5 related questions to explore further:

1. How is the Law of Sines applied to non-right triangles?
2. Can the Law of Cosines be used for this type of problem?
3. What happens if one of the angles is obtuse in a triangle like this?
4. How do changes in the angles affect the distances to the sailboat?
5. Could this problem be solved without trigonometry?

Tip: Always ensure that the sum of the angles in a triangle adds up to $180^\circ$ to verify if your angle calculations are correct.