A submarine is moving parallel to the surface of the ocean at a depth of 626 m. It begins a constant ascent so that it will reach the surface after travelling a distance of 4420 m.
a) What angle of ascent, to the nearest tenth of a degree, did the submarine make? (3
marks)
b) How far did the submarine travel horizontally, to the nearest metre, during its ascent to
the surface? (3 marks)

a) the angle of ascent is 8.2°

b) the horizontal distance traveled is 4375 m

Step-by-step explanation:

depth of ocean = 626 m

distance traveled in the ascent = 4420 m

This is an angle of elevation problem with

opposite side to the angle = 626 m

hypotenuse side = 4420 m

a) angle of ascent ∅ is gotten from

sin ∅ = opp/hyp = 626/4420

sin ∅ = 0.142

∅ = 0.142

∅ = 8.2°  this is the angle of ascent of the submarine.

b) The horizontal distance traveled will be gotten from Pythagoras theorem

= +

The horizontal distance traveled will be the adjacent side of the right angle triangle formed by these distances

= +

adj =

adj = 4375 m  this is the horizontal distance traveled.

answer: the answer to your question is parallel

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide