The official visual flight rules require that the distance between the ground and the bottom of the clouds to be more than 1000 feet for a plane to fly without instrumentation. at night, cloud height can be determined by aiming a searchlight straight upward and having an observer find the angle of elevation to the point at which the light hits the clouds. suppose the observer stands 800 feet away from the searchlight and sights the beam of light on the clouds with an angle of elevation 51 degree. what is the height of the cloud cover? feet would the visual flight rules allow a plane to fly without instrumentation in these conditions? answer 'yes' or 'no'.

height  =  h  = 987.92

Step-by-step explanation:  See Annex

The official Visual Flight Rules require more than 1000 ft , as a distance between the ground and bottom of clouds, for a plane to fly without instrumentation

We have to determine,  with available data if we are in position of that distance

If we look at picture (in annex ) tangent of  the elevation angle 51⁰  is

from tables   tan 51⁰  = 1.2349    and we know tan ∠α  = opposite leg/ adjacent leg

then

tan 51⁰    =  1.2349  =  h/ 800

h  =  987.92

So we don´t have enough height to let flying without instrumentation

300

step-by-step explanation:

28-600w < 300

28-600w > 300

600-28w < 300

600-28w > 300

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle   we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.