I’m assuming ‘dives further’ means to go directly down
the angle of elevation of the ship from the submarine is equal to the angle of depression of the submarine from the ship, if we assume the sea level is perpendicular to ‘directly down’.
let both of these angles to be = $ when the submarine is at A and ¥ when the submarine is at B (excuse the lack of easily accessible variables as keys)
then this become a simple trig problem:
A)
Let O be the position of of the ship, and C be the original position of the submarine.
therefore, not considering direction
|OC| = 1.78km = 1780m
|CA| = 45m
these are the adjacent and opposite sides of a right angled triangle.
But tan($) = opp/adj = |CA|/|OC| = 45/1780
therefore $ = arctan(45/1780) which is roughly 1.45 degrees,
B)
similarly, noting that |CB| = |CA| + |AB| = 45 + 62 = 107m
tan(¥) = 107/1780
¥ = arctan(107/1780) which is roughly 3.44 degrees
Answer:
your answer is 3 this is a simple equation type this into to brainly next time it will give you the answer
28 might be a parallelogram
.0091 rounded to the nearest thousandths would be .009
Answer: 5 cookies
Step-by-step explanation: 8 + .4c = 2.5 + 1.5c
8 + 0.4c = 2.5 + 1.5c
8 · 10 + 0.4c · 10 = 2.5 · 10 + 1.5c · 10
80 + 4c = 25 + 15c
4c = 15c - 55
4c - 15c = 15c - 55 - 15c
-11c = -55
c = 5
