A13x2+390x here is the answer
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:
52
52
is the correct answer the probability that card drawn is a faace card
There's many properties you can use to find an unknown angle.
There are too many to lists but one core example would be an isosceles triangle that has two adjacent sides and angles.
Let's say that the sides of an isosceles triangle are any number "x"
now since two sides of the triangle are the same we can add these two x's together.
x+x = 2x
now the other side of the triangle can be anything you like. We can call it 4x for this example.
now if we add them all together we'll get 4x+2x=6x
Now since the angles of a triangle add up to 180 degrees
we can equate 6x=180 leaving x to be 30.
Now since x belongs to both sides of the triangle we can say that both angles are congruent as well because the two sides of the triangle are congruent. This is a known triangle law.
Since both angles are now 30 degrees this will leave us with 2(30) = 60
now if we subtract 180 - 60 we'll get 120 which is the remainder of the 3rd angle of the side that corresponds with 4x.
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