Initially a ship and a submarine are stationary at sea level, positioned 1.78km apart. The submarine then manoeuvres to position
A, 45m directly below its starting point. In a second manoeuvre, the submarine dives a further 62m to position point B. A) find the angle of elevation of the ship from the submarine is at position A.
B) find the angle of elevation of the ship when the submarine is at position B.
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,
234/9 equals 26, that means there would be 26 pieces of candy in 9 bags, therefore 26x5 equals 130, and just to make sure, divide 130/26 to check if its wrong, but it should be 130. :) hope this helps you out!
