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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Thursday, February 21, 2008 at 11:21pm.

An airplane takes off from the ground and reaches a height of 500 feet after flying 2 miles. Given the formula H=dtan times theta, where H is the height of the plane and d is the distance (along the ground) the plane has flown, find the angle of ascent theta at which the plane took off.

- Trigonometry -
**drwls**, Friday, February 22, 2008 at 1:18amH = d tan0

tan0 = H/d = 500 ft/10,560 ft = .04735

0 = 2.71 degrees

- Trigonometry -
**drwls**, Friday, February 22, 2008 at 1:19am"0" was supposed to be the symbol theta in my previous post

- Trigonometry -
**Guido**, Friday, February 22, 2008 at 8:44amI want to expand drwls' reply.

First, convert 2 miles to feet.

If 1 mile = 5280, then 2 miles = 10,560 feet.

Secondly, you must re-arrange the given formula because we are looking for an angle NOT height or distance on ground.

Let t = theta FOR SHORT.

tan(t) = 500ft/2 miles

tan(t) = 500ft/10,560ft

tan(t) = 0.047348485

Since we are looking an an angle, you must use the tangent invere key on your calculator. By the way, you must ALWAYS use the inverse key when searching for a trig function. I will do this in just a minute.

Before dealing with the angle itself, the decimal number 0.047348485 must be rounded to the nearest hundred- thousandths; it becomes 0.04735.

We now have:

tan(t) = 0.04735 We take the tangent inverse of the right side of the equation.

NOTE: tan^-1 is read: "tangent inverse" NOT tangent to the negative one. Is this clear?

t = tan^-1(0.04735)

t = 2.71 degrees

NOTE: I rounded 2.710843763 to the nearest hundreths and the answer became 2.71 degrees. This is what drwls did without going into detail.

I think detail is vital when tutoring online.

Done!

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