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:
the answer is 0
Step-by-step explanation:
(1/3)(21)-1-(1/2)(12)
The first thing you do is that you multiply 1/3 times 21/1 which equal 7 then you multiply 1/2 times 12/1 which that equal 6 then you solve 7-1-6 and you get 6-6 which equal 0 i hope this help you and i hope you have a goood day.
Answer:
C=84
Step-by-step explanation:
Since you use 3 for pi. You plug the radius into the formula. C=2×3x14=84
Answer:
2{3[9+4(7-5)-4]}
2{3[9+4(2)-4]}
2{3[13(2)-4]}
2{3[26-4]}
2{3[22]}
2{66}
132
Step-by-step explanation:
