Question

Juan and Romella are standing at the seashore 10 miles apart. The coastline is a straight line between them. Both can see the same ship in the water. The angle between the coastline and the line between the ship and Juan is 35 degrees. The angle between the coastline and the line between the ship and Romella is 45 degrees. How far is the ship from Juan?

Answer 1

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.

$\frac{10}{sin 100} = \frac{x}{sin 45}$

By trignometric ratios,

$sin 45 = \frac{1}{\sqrt{2}}$

$sin 100 = 0.9848$

Substituting the values we get,

$\frac{10}{0.9848}=\frac{x}{\frac{1}{\sqrt{2}}}$

$x=\frac{10 \times 1}{0.9848 \times \sqrt{2}}=7.18$

Thus distance between ship and Juan is 7.18 miles approximately