xand y are light house. y being 20km due east of x and from a ship due south of x the bearing of y was 055°. what is the distance of the ship from y and the distance of the ship from x?

I) |xz| ≈ 28.6 km

II) |yz| ≈ 34.8 km

Step-by-step explanation:

Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)

|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°

Using Trigonometric ratio - SOHCAHTOA

I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20

|xz| = 20 * Tan 55 = 20 * 1.428

|xz| = 28.56 km

|xz| ≈ 28.6 km

II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|

|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°

|yz| = 20 ÷ 0.574 = 34.84 km

|yz| ≈ 34.8 km

two data sets with the same median will not have the same varaibility

step-by-step explanation: