Question

1. a cargo ship is 4.2 miles from a lighthouse, and a fishing boat is 5.0 miles from the lighthouse, as shown below. the angle between the straight lines from the lighthouse to the 2 vessels is 5°. the approximate distance, in miles, from the cargo ship to the fishing boat is given by which of the following expressions? (note: the law of cosines states that for any triangle with vertices a, b, and c and the sides opposite those vertices with lengths a, b, and c, respectively, c² = a² + b² - 2ab cos c.) a. √((5.0)²-(4.2)²) b. √((4.2)²+(5.0)² - 2·4.2·5.0 cos 5°) c. √((4.2)²+(5.0)² + 2·4.2·5.0 cos 5°) d. √((4.2)²+(5.0)² - 2·4.2·5.0 cos 85°) e. √((4.2)²+(5.0)² + 2·4.2·5.0 cos 85°)

Explanation:

Step1: Identify the sides and angle in the triangle

Let $a = 4.2$, $b = 5.0$, and $C=5^{\circ}$. We want to find the length of the side $c$ (distance between the cargo - ship and the fishing boat).

Step2: Apply the law of cosines formula

The law of cosines is $c^{2}=a^{2}+b^{2}-2ab\cos C$. Substituting $a = 4.2$, $b = 5.0$, and $C = 5^{\circ}$ into the formula, we get $c^{2}=(4.2)^{2}+(5.0)^{2}-2\times4.2\times5.0\cos5^{\circ}$.

Step3: Solve for $c$

Taking the square - root of both sides to find $c$, we have $c=\sqrt{(4.2)^{2}+(5.0)^{2}-2\times4.2\times5.0\cos5^{\circ}}$.

Answer:

B. $\sqrt{(4.2)^{2}+(5.0)^{2}-2\times4.2\times5.0\cos5^{\circ}}$