There is nothing attached
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
Given :
Center of sphere , C( 1 , -1 , 6 ) .
To Find :
Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .
Solution :
a)
Distance of the point from xy-plane is :
d = 6 units .
So , equation of circle with center C and radius 6 units is :
b)
Distance of point from yz-plane is :
d = 1 unit .
So , equation of circle with center C and radius 1 units is :
c)
Distance of point from xz-plane is :
d = 1 unit .
So , equation of circle with center C and radius 1 units is :
Hence , this is the required solution .
A line that is tangent to each of the two coplanar circles
