Question about Super Tutor Trigonometry (ESDTRIG) for PC

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Posted on Jan 02, 2017

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It's quite permissible to do so, however a general aviation pilot in a low performance plane should be ready for fast instructions and quite a bit of maneuvering to stay out of the way. The best time to do it would be at night. Here's a youtube video of one doing it. https://www.youtube.com/watch?v=KKvWn317tpU

Jan 04, 2017 | Aircrafts

It is very common for small airports not to have a control tower, radar, or communications equipment. There are very well developed procedures for "uncontrolled airports". There is a standard traffic pattern that aircraft fly at almost all airports (consisting of a downwind, base, final, and upwind leg) and there are specific radio calls that are supposed to be made at certain points in the pattern. Most airports have a fixed base operator to supply fuel and services. They often monitor the common traffic frequency and supply some info to pilots about wind direction and runway in use. There's also a specific way to enter the pattern - usually at a 45 degree of the downwind leg, Yes it's possible for aircraft to collide and it happens several times a year - usually when a low wing airplane is above a high wing airplane in the pattern. Neither can see the other so occasionally that can happen.

Jan 04, 2017 | Aircrafts

You're stuck with a water landing (ditching). Hopefully at altitude the pilot can see a ship and make the landing in the vicinity of the ship for quick rescue. It so rare for a plane to lose power to all engines though that you should never hear of such an incident. Depending on the route, however, Bermuda may be within gliding distance.

Jan 04, 2017 | Aircrafts

As long as the plane is equipped for icing, which all airlines are, there is not really any added danger to flying in snow. Braking on landing may be affected if there's any accumulation but they keep close track on braking ability as each plane lands at the airport. Snow won't affect the way the plane flies as long as the wings are clear of snow on takeoff.

Jan 04, 2017 | Aircrafts

Using the old (pre GPS) method you would use the sectional and plotter (aviation ruler) and measure the distance. Using a GPS you can usually turn on distance rings on the screen or if you have the airport selected as destination you can read it right off the screen. If it's your home base you should learn the landmarks and their distances from the airport. In the Miami area, there are so many airports and landmarks, if you're flying there, during your preflight preps you should measure out distances to some landmarks that you plan to pass over and mark them on the chart or flight log.

Jan 04, 2017 | Aircrafts

Students and certificated pilots both are required to remain current and to have a check ride (flight review with an instructor) every 24 calendar months. The ones you're seeing are probably going out to the practice area to practice their flight maneuvers. If a pilot doesn't practice regularly his or her skills can deteriorate rapidly.

Jan 04, 2017 | Aircrafts

Draw the airport. (X marks the spot) Draw the distances from the airport (to scale) to indicate the position of the Plane. Use a protractor to work out the correct angle.

To Calculate:

a=48 b=49 Angle= Tan-1(a/b)

To Calculate:

a=48 b=49 Angle= Tan-1(a/b)

Sep 25, 2013 | Cars & Trucks

v will represent plane's airspeed.

We can divide pilot's trip in 2 parts:

1.5-t=200/(v-30)

We also know that t=200/v so we substitute that into equation:

1.5-200/v=200/(v-30)

To solve this equation we multiply it with v*(v-30)=v^2-30v first. We get:

1.5(v^2-30v)-200(v-30)=200v

Now we get rid of brackets:

1.5v^2-45v-200v+600-200v=0

or

1.5v^2-445v+600=0

This is quadratic equation, so we get 2 solutions (sqrt= square root):

We can divide pilot's trip in 2 parts:

- first 200km, when he is flying with speed v. The time it takes him to fly this part of trip is t=200/v
- last 200km, when he is flying with speed v-30. The time it takes him to fly this part is 200/(v-30)

1.5-t=200/(v-30)

We also know that t=200/v so we substitute that into equation:

1.5-200/v=200/(v-30)

To solve this equation we multiply it with v*(v-30)=v^2-30v first. We get:

1.5(v^2-30v)-200(v-30)=200v

Now we get rid of brackets:

1.5v^2-45v-200v+600-200v=0

or

1.5v^2-445v+600=0

This is quadratic equation, so we get 2 solutions (sqrt= square root):

- v=(445+sqrt(445^2-4*1.5*600))/(2*1.5)=295.3 km/h
- v=(445-sqrt(445^2-4*1.5*600))/(2*1.5)=1.4 km/h

Sep 08, 2011 | Casio ClassPad 300 Calculator

a2+b2=c2

a being your height, b is your distance from the airport and c is the distance from the airport to the plane........ so the answer is 5.385164807134504 miles

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a being your height, b is your distance from the airport and c is the distance from the airport to the plane........ so the answer is 5.385164807134504 miles

Pls Rate this post

Oct 08, 2009 | SoftMath Algebrator - Algebra Homework...

Debs,

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.

Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311

- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:

square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.

Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311

- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:

square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

Dec 19, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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