Asked by anonymous
The angles of elevation θ and ϕ to an airplane are being continuously monitored at two observation points A and B, respectively, which are 5 miles apart, and the airplane is east of both points in the same vertical plane. (Assume that point B is east of point A.)
(a) Draw a diagram that illustrates the problem.
(b) Write an equation giving the distance d between the plane and point B in terms of θ and ϕ.
(c) Use the equation from part (b) to find the distance between the plane and point B when θ = 39° and ϕ = 60°. (Round your answer to two decimal places.)
(a) Draw a diagram that illustrates the problem.
(b) Write an equation giving the distance d between the plane and point B in terms of θ and ϕ.
(c) Use the equation from part (b) to find the distance between the plane and point B when θ = 39° and ϕ = 60°. (Round your answer to two decimal places.)
Answers
Answered by
anonymous
it is an online assignment and i really don't understand most of these assignment question that's i am post these questions here
Answered by
Damon
P is plane location
working with triangle ABP
AB= 5
angle at A = 39 = θ
angle at B = 180 - 60 = 120 = 180 - ϕ
so angle at P = 180 - 39 -120 = 21 deg
sin 21/5 = sin 39 / BP
working with triangle ABP
AB= 5
angle at A = 39 = θ
angle at B = 180 - 60 = 120 = 180 - ϕ
so angle at P = 180 - 39 -120 = 21 deg
sin 21/5 = sin 39 / BP
Answered by
anonymous
hey damon, thanks, is the answer for sin 21/5 = sin 39 / BP is 8.780 and is 8.780 is the answer for question b??
Answered by
Damon
I do not feel like running the numbers. That looks reasonable.
Answered by
anonymous
question b says i have to make an equation? how do i make that equation?
Answered by
Damon
angle B = 180 - ϕ
angle P = 180 - θ - (180 - ϕ) = ϕ - θ
so
sin (ϕ - θ)/ 5 = sin θ / d
angle P = 180 - θ - (180 - ϕ) = ϕ - θ
so
sin (ϕ - θ)/ 5 = sin θ / d
Answered by
oobleck
for (b) note that if the plane is x miles east of B at an altitude h,
h/x = tanφ
h/(x+5) = tanθ
eliminating h,
xtanφ = (x+5)tanθ
x = 5tanθ/(tanφ-tanθ)
Now note that
x/d = cosφ
d = x/cosφ = (5tanθ/(tanφ-tanθ))/cosφ
= 5tanθ/(cosφ(tanφ-tanθ))
h/x = tanφ
h/(x+5) = tanθ
eliminating h,
xtanφ = (x+5)tanθ
x = 5tanθ/(tanφ-tanθ)
Now note that
x/d = cosφ
d = x/cosφ = (5tanθ/(tanφ-tanθ))/cosφ
= 5tanθ/(cosφ(tanφ-tanθ))
Answered by
Damon
Relax, we will get there. Just keep your parachute on.
Answered by
Damon
I bet we could find a harder way if we really put our minds to it.
Answered by
Damon
I just do navigation, not mathematics and on ocean not in air. So perhaps I am lost :)
Answered by
anonymous
how about question c?
Answered by
Damon
But I did part c at the beginning
AB= 5
angle at A = 39 = θ
angle at B = 180 - 60 = 120 = 180 - ϕ
so angle at P = 180 - 39 -120 = 21 deg
sin 21/5 = sin 39 / BP
AB= 5
angle at A = 39 = θ
angle at B = 180 - 60 = 120 = 180 - ϕ
so angle at P = 180 - 39 -120 = 21 deg
sin 21/5 = sin 39 / BP
Answered by
anonymous
what's the ending of c?
Answered by
anonymous
Hey Damon, formula for question b is wrong
Answered by
Ms. Sue
Anonymous -- what is the correct formula for question b?
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