One hundred feat directly overhead, a plane and a dirigible start flying due east. If the plane travels three?

One hundred feat directly overhead, a plane and a dirigible start flying due east. If the plane travels three times faster than the dirigible, what is the greatest angle of sight between the two vehicles from your position? Assume the planes passes from sight after 6000 feet.One hundred feat directly overhead, a plane and a dirigible start flying due east. If the plane travels three?Lets assume the distance dirigible has gone x feet. Then plane will have gone 3x feet.

Also assume angle of plane from the vertical is P, and of dirigible is D. (from the vertical position)



Then tan(P) = 3x/100

and tan(D) = x/100



We need to find maxima of (P-D) = y



then y = - tan^-1 ( x/100 ) + tan^-1 ( 3x/100 )



then dy/dx = - ( 1/ (1+ (x/100)^2) * 1/100) + ( 1/ (1+ (3x/100)^2) * 3/100 )

[ using tan-1 derivative formula's ]



At maxima, the derivative is zero. So equating %26amp; rearranging, we get --



3+ 3* (x/100)^2 = 1 + (3x/100)^2



solving the above --



2 + 3*x^2 /10000 = 9* x^2/ 10000

OR re-arranging again --



3* x^2 / 10000 = 1

x^2 = 3333.33

x = 57.73 ft for the maximum angle.

So dirigible travelled 57.73 ft, and plane travelled 173.19 ft

The 2 angles are -- ~ 30 deg and 60 deg.



So Max angle is ( 60 - 30 ) = 30 degrees.One hundred feat directly overhead, a plane and a dirigible start flying due east. If the plane travels three?They're directly overhead to start with, and at the furthest point the furthest vehicle will be 6000 feet away (accursed Americans with their imperial foot%26amp;inch system!)



It will be the plane reaching the 6000ft marker first, as it is travelling faster.

Speed of plane = 3 * speed of dirigible

Distance covered by plane = 6000

Distance covered by dirigible = 6000 / 3 = 2000

So the plane will be 6000ft from you, and the dirigible only 2000 feet away.



I assume they remain at 100ft above you.



So now you have a scalene triangle of height 100 and horizontal distance (6000-2000)=4000ft and you want the angle opposite that 4000ft side (the angle of sight between the two vehicles.

You also have a right angled triangle height 100 and horizontal distance 6000ft,

and another right angled triangle height 100 and horizontal distance 2000ft,

where all three triangles overlap each other.



The angle made from you to the dirigible is given by tanD = 2000 / 100 = 20

The angle made from you to the plane is given by tanP = 6000 / 100 = 60

The angle you're looking for is the difference between these two angles, so...

P - D = inverse tan(60) - inverse tan(20) = 89.045... - 87.137... = 1.907...



So the greatest angle of sight is 1.9 degrees (1dp)

yamaha atv