00:01
In this problem, we have been given that one of the airplane is at an altitude of 132 and 11 .8 kilometers along the horizontal direction with respect to the tower at 32 .8 degree south of east.
00:21
So here, let's say this point, the origin that's representing the position of the tower, let's say this is t.
00:28
And let's indicate east, south, west, and north directions.
00:38
So as we are given that, the aeroplane, let's say the first airplane is a, that's at an angle of 32 .8 degree south of west.
00:50
So basically it will be in this direction, and the angle here is 32 .8 degree.
00:58
So the horizontal distance is 11 .8 kilometers, so that's 11 .18 into 1000.
01:12
So that comes up to be 118 -00 meters.
01:15
So here we take this as the x -axis, this as the y -axis, and perpendicular to this coming out of the screen, we take that as the x -exis.
01:27
So we can have the position coordinate of this plane.
01:34
So that position coordinate since it is at 11 .8 kilometers, so its x and y coordinate can be obtained by taking its component along x and y directions.
01:48
So along x direction, its component will be 11 .8 cost 32 .8 degree, and along y direction it will be 11 .8.
01:59
Sine 32 .8 degree.
02:04
So now we get the position coordinate.
02:07
So first we figure out the value of cost 32 .8 and we multiply that with 11 .8.
02:16
That comes out to be 9 .92.
02:21
And again we multiply sine 32 .8 with 11 .8.
02:30
So that comes out to be 6 .4 approximately.
02:35
So we get the position coordinate of this aeroplane, that is minus 9 .92, minus 6 .4, and 11 ,800 meters.
02:54
So this we have taken in terms of kilometers only.
02:58
And along xx, that's the altitude here.
03:01
That's 1132.
03:05
So we'll take even that in kilometers, so it will be 1 .13 kilometers approximately.
03:12
So this is the position coordinate of plane a and another plane that's at an altitude of 5257 meters.
03:22
So let's take that as the plane b.
03:25
So altitude we represent as the z coordinate...