A plane flies horizontally at an altitude of 6 km and passes directly over a tracking telescope on the ground. When the angle of elevation is /3, this angle is decreasing at a rate of /3 radians per minute. How fast is the plane traveling at that time?

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Step-by-step explanation:

We have to make a few assumptions first. We assume the angle provided (12.9 degrees) is the plane's attitude, that is, the angle with respect to the horizontal. We also assume the plane's nose is angled toward the ground because you asked for rate of descent (not dissent). We also assume that all of the plane's airspeed is in the same direction as its attitude (which is often not the case in actual flight, where an aircraft's velocity vector may have a different angle than the angle the plane makes with the ground). So, we have an airplane that's traveling at 278 km per hour at an angle that is -12.9 degrees to the horizontal. (The minus sign indicates a downward direction.) That describes a vector having vertical and horizontal components. We are interested in the vertical component, which is given by the formula Vy = Vsin(theta) = 278sin(-12.9) = -62.1. The rate of descent, therefore, is 62.1 km/hr. It is not necessary to use the minus sign if you say "rate of descent," because the minus sign is implied. If you are asked what the vertical velocity is, then you must use the minus sign to indicate whether it's positive (ascending) or negative (descending).

whats the

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72