�Two radar stations A and B, are separated by the distance a = 500 m, track the plane C by recording the angles α and β at 1-second intervals. If three successive readings

are
t (s) 9 10 11
α 54.80◦ 54.06◦ 53.34◦
β 65.59◦ 64.59◦ 63.62◦
calculate the speed v of the plane and the climb angle γ at t = 10 s.
The coordinates of the plane can be shown to be,

x = a*((tan β)/(tan β − tan α))

y = a*((tan α tan β)/(tan β − tan α))

maybe I'm missing something, but if angles α and β are in the x-y plane, how do we figure the altitude of the plane, and hence any climb rate? No mention is made of altitude, either of the radar stations, or of the plane.

No figure here, so I'm not sure of the 3-D layout.

nikaw

To calculate the speed of the plane (v) and the climb angle (γ) at t = 10 seconds, we can use the coordinate equations provided:

x = a * (tan β / (tan β - tan α))
y = a * (tan α * tan β / (tan β - tan α))

Let's start by substituting the values at t = 10 seconds into these equations:

t = 10 s
α = 54.06°
β = 64.59°

x = a * (tan β / (tan β - tan α))
= 500 * (tan(64.59°) / (tan(64.59°) - tan(54.06°)))

To calculate this, we need to convert the angles from degrees to radians since trigonometric functions typically take radians as input. We can use the formula:

radians = degrees * (π / 180)

Using this formula, we can convert our angles to radians:

α_rad = 54.06° * (π / 180) ≈ 0.9437 radians
β_rad = 64.59° * (π / 180) ≈ 1.1265 radians

Substituting these values into the equation for x:

x = 500 * (tan(1.1265) / (tan(1.1265) - tan(0.9437)))

Now we can use a calculator or computer program to evaluate the expression for x.

Once we have the value of x, we can calculate y using the equation:

y = a * (tan α * tan β / (tan β - tan α))

Substituting the values at t = 10 seconds into this equation:

y = 500 * (tan(0.9437) * tan(1.1265) / (tan(1.1265) - tan(0.9437)))

Again, we can use a calculator or computer program to evaluate the expression for y.

Finally, to calculate the speed (v) and the climb angle (γ) at t = 10 seconds, we need to find the time derivative of x and y:

v = sqrt(dx/dt^2 + dy/dt^2)
γ = atan(dy/dt / dx/dt)

To find the derivatives, we can use numerical differentiation with the given data points at t = 9 s and t = 11 s. The time interval is 1 second, so the derivatives can be approximated by:

dx/dt ≈ (x(t=11) - x(t=9)) / 2
dy/dt ≈ (y(t=11) - y(t=9)) / 2

Substituting the values of x and y at t = 9 s and t = 11 s, we can calculate the derivatives.

Once we have the derivatives, we can calculate v and γ using the formulas above.

Note: Make sure to use appropriate units for the angles (degrees or radians) when performing calculations and interpreting the results.