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
Divide the long leg by the square root of 3 to find the short leg. Double that figure to find the hypotenuse.
Answer:
108 degrees.
Step-by-step explanation:
2x + 16 = 3x - 12 (corresponding angles; AB parallel to CD).
16 + 12 = 3x - 2x
x = 28.
So the angle marked 3x - 12 = 3(28) - 12
= 72 degrees.
Angle y is adjacent to the angle so
y = 180 - 72
= 108 degrees.
Answer:
5 5/7 acres
Step-by-step explanation:
The product is ...
(2 1/7 acres/section)(2 2/3 sections) = (15/7)(8/3) acres = 40/7 acres
= 5 5/7 acres
The farmer owns 5 5/7 acres of land.
Answer: 
Step-by-step explanation:
For this exercise you need to remember:
1) The mulitplication of signs:
2) By definition, like terms contain the same variables with the same exponent.
Then knowing this, and given the equations:
and 
You must subtract the like terms, which means that you must subtract the numerical coefficients of each term.
Then, you get the result is:
