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

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this