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
Six tenths. Hope that helps
You just need to multiply 185 by 3. 185 x 3 = 555
Answer:
A = 23.6 units^2
Step-by-step explanation:
Let the base be 8 (as shown).
Find the height of the triangle: Find the supplement of the 100 degree angle; it is (180 - 100), or 80 degrees. The side "opposite" this 80-degree angle is the height of the triangle:
height
sin 80 degrees = ---------------
6
and so the height of the triangle is h = 6 sin 80 degrees, or 5.91 units.
The area of the triangle is found using A = (1/2)(base)(height), which here amounts to:
A = (1/2)(8 units)(5.91 units), or
A = 23.64 units^2, or A = 23.6 units^2
Answer:
m∠60
Step-by-step explanation:
30 + 90 = 120
180-120=60
