Boat's Direction Change Calculation Using Law Of Cosines

Ahoy, math enthusiasts! Let's dive into a fascinating nautical problem that combines geometry, trigonometry, and a touch of adventure. We're going to chart the course of a boat, calculate distances, and uncover the mystery of its directional change. Get ready to set sail on this mathematical voyage!

The Boat's Tale: A Journey North and Southwest

Imagine a boat embarking on a journey. This vessel initially travels 28 miles due north, a straight and unwavering path. Then, the captain makes a turn, veering $x^{\circ}$ southwest. After this directional shift, the boat cruises for another 25 miles before dropping anchor. The crucial piece of information we have is that, upon stopping, the boat finds itself 18 miles away from its original starting point. This sets the stage for our mathematical exploration.

To truly grasp the situation, let's visualize this journey. Picture a triangle forming in the water. The first leg, 28 miles north, forms one side. The second leg, 25 miles southwest, creates another side. And the direct line connecting the starting point to the final stopping point, 18 miles, closes the triangle. Our mission is to determine the angle $x$, the degree of the boat's directional change. This is where the Law of Cosines, a powerful tool in trigonometry, comes into play. Guys, this is where it gets really interesting!

Applying the Law of Cosines: Unveiling the Angle

The Law of Cosines is our key to unlocking the value of $x$. This law relates the sides and angles of a triangle, providing a formula to calculate unknown quantities. In our case, we know the lengths of all three sides of the triangle: 28 miles, 25 miles, and 18 miles. We want to find the angle opposite the 25-mile side, which is the angle formed at the starting point between the northward path and the direct line to the stopping point. Let's call this angle $\theta$ (theta).

The Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab \cos(C)$

Where:

• $c$ is the side opposite angle $C$
• $a$ and $b$ are the other two sides
• $C$ is the angle opposite side $c$

In our scenario:

• $c = 25$ miles
• $a = 28$ miles
• $b = 18$ miles
• $C = \theta$

Plugging these values into the Law of Cosines, we get:

$25^2 = 28^2 + 18^2 - 2(28)(18) \cos(\theta)$

Now, let's simplify and solve for $\cos(\theta)$:

$625 = 784 + 324 - 1008 \cos(\theta)$

$625 = 1108 - 1008 \cos(\theta)$

$1008 \cos(\theta) = 1108 - 625$

$1008 \cos(\theta) = 483$

$\cos(\theta) = \frac{483}{1008}$

$\cos(\theta) \approx 0.4791666667$

To find $\theta$, we take the inverse cosine (arccos) of 0.4791666667:

$\theta = \arccos(0.4791666667)$

$\theta \approx 61.35^{\circ}$

So, the angle $\theta$ at the starting point is approximately 61.35 degrees. But hold on, we're not quite there yet! We need to find $x$, the angle of the boat's directional change southwest.

Decoding the Directional Change: Finding x

Remember that the boat initially traveled due north. A southwest direction implies an angle between 180 degrees and 270 degrees from the north direction. To find $x$, we need to consider the geometry of the situation. The angle $x$ represents the change in direction from north to southwest. This is a critical step, guys, so pay close attention!

Imagine a compass rose. North is at 0 degrees, East is at 90 degrees, South is at 180 degrees, and West is at 270 degrees. Southwest lies exactly between South and West, at 225 degrees. However, we need to account for the angle $\theta$ we just calculated. $\theta$ represents the angle between the northward path and the direct line to the stopping point. To find the boat's actual change in direction ($x$), we need to work with the angles formed within our triangle and the compass directions.

The direction southwest implies an angle of 225 degrees from North. However, the angle $x$ is not simply 225 degrees minus $\theta$. We need to find the interior angle of the triangle at the starting point that corresponds to the change in direction. Let's call this interior angle $\alpha$ (alpha). The angle $\alpha$ is supplementary to the angle between the northward path and the southwestern path. In simpler terms, $\alpha$ and the angle of the southwest turn add up to 180 degrees.

To visualize this, imagine extending the northward path beyond the starting point. This creates a straight line, and the angle between this extended line and the southwestern path is supplementary to $\alpha$. Since the boat turned southwest, this supplementary angle is $x$. Therefore, we have:

$\alpha + x = 180^{\circ}$

Now, consider the angle between the North direction and the direction South of West. Since Southwest is 45 degrees from South (225 degrees - 180 degrees), we can calculate the angle $\alpha$ as:

$\alpha = 180^{\circ} - (180^{\circ} - 45^{\circ}) - \theta $

$\alpha = 180^{\circ} - 135^{\circ} - 61.35^{\circ}$

$\alpha = -16.35^{\circ}$

However, this result doesn't make sense because the angle $\alpha$ cant be negative. Let's correct our approach. We need to consider the angle between the North direction and the boat's path after turning southwest. Since the boat turned $x$ degrees southwest, the angle between the North direction and the boat's path is $180 - 45 = 135$ degrees, where 45 degrees is the angle between South and Southwest.

To find $x$, the degrees the direction of the boat changed, we subtract $\theta$ from the angle representing the southwest direction relative to north. We know that southwest is 45 degrees past south, or 225 degrees from north.

However, because our reference point is the northward direction, we need to consider the interior angle within our triangle. The angle $x$ we are seeking is supplementary to the angle formed by extending the northward path and the southwestern path. Let's revisit the geometry and consider the angles more carefully.

The key insight is that the angle $x$ represents the change in direction from north. The boat turned southwest, which is generally understood as 225 degrees from north in a compass bearing. But we need to relate this to the angles we've calculated in our triangle. Let's go back to the Law of Cosines and our calculated angle $\theta$.

We know $\theta$ is approximately 61.35 degrees. This angle is crucial because it helps us determine the internal angles of the triangle. To find the angle $x$, we need to consider how the southwest direction relates to $\theta$. The boat turned southwest, but not a perfect southwest. The 18 miles displacement from the start means the turn wasn't a clean 225-degree bearing.

Let's use the Law of Sines to find another angle in the triangle. The Law of Sines states:

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

Where $a$, $b$, and $c$ are the sides of the triangle, and $A$, $B$, and $C$ are the opposite angles.

We know:

• $a = 28$ miles, $A$ is opposite the 28-mile side
• $b = 25$ miles, $B = \theta \approx 61.35^{\circ}$
• $c = 18$ miles, $C$ is opposite the 18-mile side

Let's find angle $A$:

$\frac{28}{\sin(A)} = \frac{25}{\sin(61.35^{\circ})}$

$\sin(A) = \frac{28 \sin(61.35^{\circ})}{25}$

$\sin(A) \approx \frac{28 \times 0.8771}{25}$

$\sin(A) \approx 0.9818$

$A = \arcsin(0.9818)$

$A \approx 79.02^{\circ}$

Now we can find angle $C$:

$C = 180^{\circ} - A - B$

$C = 180^{\circ} - 79.02^{\circ} - 61.35^{\circ}$

$C \approx 39.63^{\circ}$

The angle $x$ represents the degrees the boat turned from its northward path. It's related to the interior angle of the triangle at the starting point (angle $C$). Since the turn is southwest, we need to find the difference between the direction of a perfect southwest turn (225 degrees from north) and the actual turn. However, relating this directly to angle $C$ (39.63 degrees) is tricky. Let's try a different approach.

The change in direction $x$ is the exterior angle to angle $C$ in the triangle formed at the starting point. We are looking for how much the boat deviated from going straight North before turning Southwest. The angle of turn can be viewed as the supplementary angle to angle $C$ if we consider the angle formed by extending the initial Northward path.

Therefore: $x \approx 180 - 39.63 \approx 140.37$

So the change in direction is approximately 140 degrees.

Conclusion: Charting the Course to the Solution

We've navigated a complex problem involving a boat's journey, directional changes, and the application of trigonometric principles. By employing the Law of Cosines and the Law of Sines, we were able to dissect the triangle formed by the boat's path and determine the angles involved. The final answer, the degree by which the boat's direction changed, is approximately 140 degrees. This problem showcases the power of mathematical tools in solving real-world scenarios. Keep exploring, guys, and happy sailing through the seas of mathematics!