Question

An island is located 44 miles N​29​°45'W of a city. A freighter in distress radios its position as N11°19'E of the island and N15°10​'W of the city. How far is the freighter from the​ city?

Answer 1

The distance between the freighter from the​ city will be 73.71 miles.

What is law of sines?

For any triangle ABC, with side measures |BC| = a. |AC| = b. |AB| = c,

we have, by law of sines,

$\rm \dfrac{sin\angle A}{a} = \dfrac{sin\angle B}{b} = \dfrac{sin\angle C}{c}$

An island is located 44 miles, N​29​°45'W of a city.

A freighter in distress radios its position as N11°19'E of the island and N15°10​'W of the city.

Then the value of angle x will be

x = 11°19' + 15°10'

x = 26°29'

Then the value of y will be

26°29' + 29°45' + y = 180°

                             y = 123°46'

Then the distance between the freighter from the​ city will be

sin 29°45' / 44 = sin 123°46' / d

                     d = 73.71 miles

Learn more about law of sines here:

https://brainly.com/question/17289163

#SPJ1