Distance between ship and Juan is 7.18 miles approximately
Solution:
Given that Juan and Romella are standing at the seashore 10 miles apart
The coastline is a straight line between them
Using the above information, we can figure a triangle
The figure is attached below
In the triangle ABC,
A = Position of ship
B = Position of Juan
C = Position of romella
Given that the angle between the coastline and the line between the ship and Juan is 35 degrees
Angle B = 35 degrees
Given that the angle between the coastline and the line between the ship and Romella is 45 degrees
Angle C = 45 degrees
Let us first find the angle C
By angle sum property,
Angle sum property states that the angles of a triangle always add up to 180 degrees
So for triangle ABC,
angle A + angle B + angle C = 180
angle A + 35 + 45 = 180
angle A = 180 - 35 - 45
angle A = 100 degrees
To find: Distance between ship and Juan
Let "x" be the distance between Juan and ship
Using law of sines,
It states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.
[tex]\frac{10}{sin 100} = \frac{x}{sin 45}[/tex]
By trignometric ratios,
[tex]sin 45 = \frac{1}{\sqrt{2}}[/tex]
[tex]sin 100 = 0.9848[/tex]
Substituting the values we get,
[tex]\frac{10}{0.9848}=\frac{x}{\frac{1}{\sqrt{2}}}[/tex]
[tex]x=\frac{10 \times 1}{0.9848 \times \sqrt{2}}=7.18[/tex]
Thus distance between ship and Juan is 7.18 miles approximately
