Question

1. rounded places please;4. calculator okcalculator ready formshow work as calculator ready formderive the law of cosinesexplain when and why the law of sines can produce two triangles.2. the beach communities of san clemente(s) and long beach(l) are 41 miles apart, along a fairly straight stretch of coastline. shown in the figure is the triangle formed by the two cities and the town of avalon(a) at the south east corner of santa catalina island. angles als and asl are found to be 60° and 47.2° respectively.a. find the distance from avalon to each of the two cities.b. find the bearing from avalon to long beach. assume that the bearing from san clemente to long beach is n90°w

Explanation:

Step1: Find angle at Avalon

First, calculate $\angle LAS$ using the angle sum of a triangle:
$$\angle LAS = 180^\circ - 60^\circ - 47^\circ2' = 72^\circ58'$$
Convert $47^\circ2'$ and $72^\circ58'$ to decimal degrees:
$47^\circ2' = 47 + \frac{2}{60} \approx 47.0333^\circ$
$72^\circ58' = 72 + \frac{58}{60} \approx 72.9667^\circ$

Step2: Apply Law of Sines for AL

Let $LS = 41$ miles, $AL = x$, $AS = y$. Use Law of Sines:
$$\frac{x}{\sin(47.0333^\circ)} = \frac{41}{\sin(72.9667^\circ)}$$
$$x = \frac{41 \times \sin(47.0333^\circ)}{\sin(72.9667^\circ)}$$
$\sin(47.0333^\circ) \approx 0.7314$, $\sin(72.9667^\circ) \approx 0.9563$
$$x \approx \frac{41 \times 0.7314}{0.9563} \approx 31.5 \text{ miles}$$

Step3: Apply Law of Sines for AS

$$\frac{y}{\sin(60^\circ)} = \frac{41}{\sin(72.9667^\circ)}$$
$$y = \frac{41 \times \sin(60^\circ)}{\sin(72.9667^\circ)}$$
$\sin(60^\circ) \approx 0.8660$
$$y \approx \frac{41 \times 0.8660}{0.9563} \approx 36.9 \text{ miles}$$

Step4: Calculate bearing for part b

The bearing from San Clemente to Long Beach is N90°E (east along latitude). From point A, the angle between north and line AL is:
$90^\circ - 60^\circ = 30^\circ$
So the bearing is N30°E.

Answer:

a. Distance from Avalon to Long Beach (AL) ≈ 31.5 miles; Distance from Avalon to San Clemente (AS) ≈ 36.9 miles
b. Bearing from Avalon to Long Beach: N30°E