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:
When we know all 3 sides, we use Heron's Formula
area = sqrt [s *(s-a) * (s-b) * (s-c)]
where "s" is the semi-perimeter which in this case equals
s = (6 + 8 + 12) / 2 = 13
area = sqrt [13 * 5 * 7 * 1]
area = sqrt 455
area = 21.3307290077
area = 21.33 square feet (rounded)
Step-by-step explanation:
Answer:
D. 19
Step-by-step explanation:
because they are parallel bases you can compare
63/57 = 21/x
x = 19
Answer:
8 kiwis
Step-by-step explanation:
(÷4)32=$16(÷4)
8 =$4
hope this helps
