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.)
Write an equation giving the distance d between the plane and point B in terms of θ and ϕ.
is someone going to help me? i need help
I answered this
angle B = 180 - ϕ
angle P = 180 - θ - (180 - ϕ) = ϕ - θ
so
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sin (ϕ - θ)/ 5 = sin θ / d
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Since you didn't like my other formula, let's try this:
If the plane is at height h, then
h = 5/(cotθ-cotϕ)
d = h/sinϕ = 5/(sinϕ(cotθ-cotϕ))
Equivalent trig expressions can be massaged to appear very different
Well, well, well, looks like we have a bit of an observation situation here. Have no fear, Clown Bot is here to come up with an equation that'll solve this mystery!
Now, from what I understand, we have two angles of elevation, θ and ϕ, and we need to find an equation that relates the distance between the plane and point B, d, to these angles. Ready? Let's go!
First, let's create a right triangle. We know that the plane is east of both points A and B, which means that the vertical distance between the plane and point A is the same as the vertical distance between the plane and point B. Let's call this distance h.
So, in right triangle A, the opposite side would be h, and the adjacent side would be 5 miles (the distance between A and B). The angle θ is the angle of elevation from point A, which means that the tangent of θ is equal to h divided by 5. Got it?
Similarly, in right triangle B, the opposite side would also be h, and the adjacent side would be 0 (since the plane is directly above point B). The angle ϕ is the angle of elevation from point B, which means that the tangent of ϕ is equal to h divided by 0. Wait a minute... we can't divide by 0! Clown Bot's not a magician!
But fear not, my friend! Since we know that the plane is directly above point B, the distance d between the plane and point B is equal to h. So, we can rewrite our equation as follows:
tan(θ) = d / 5
Now we have an equation giving the distance d between the plane and point B in terms of θ. Not too shabby, eh?
To find an equation that relates the distance d between the plane and point B to the angles of elevation θ and ϕ, we can use trigonometry. Let's start by visualizing the scenario:
We have a right triangle formed by the line AB (5 miles), the line BC (distance d), and the line AC (distance between the plane and point A). Here, point C represents the plane.
Now, let's consider the angles of elevation. The angle θ is the angle between the line AB and the line AC, while the angle ϕ is the angle between the line AB and the line BC.
Since θ and ϕ are angles of elevation, they relate to the triangle angles. Specifically, θ is the complement of angle ABC (which we'll call α), and ϕ is the complement of angle ACB (which we'll call β).
Now, we can apply trigonometry to the triangle ABC using the tangent function. The tangent of an angle is defined as the opposite side divided by the adjacent side.
In this case, we can consider:
tan(α) = AC / AB, and
tan(β) = BC / AB.
Let's isolate BC in the equations:
BC = AB * tan(β),
AC = AB * tan(α).
Next, let's substitute the values for α and β:
α = 90° - θ,
β = 90° - ϕ.
Substituting these values into the equations, we get:
BC = AB * tan(90° - ϕ),
AC = AB * tan(90° - θ).
Simplifying further:
BC = AB * cot(ϕ),
AC = AB * cot(θ).
Finally, we can substitute AB with 5 miles:
BC = 5 miles * cot(ϕ),
AC = 5 miles * cot(θ).
Therefore, the equation giving the distance d between the plane and point B in terms of θ and ϕ is:
d = 5 miles * cot(ϕ).