Applications of TrigonometryDate: 07/19/98 at 19:13:16 From: Susie Lee Subject: Trig I have three questions. 1) A pilot is flying at an air speed of 241 mph in a wind blowing 20.4 mph from the east. In what direction must the pilot head in order to fly due north? What is the pilot's speed relative to the ground? 2) Find the area of the following triangle: A = 115 deg 20', b = 9.23 m, c = 12.1 m. I know how to find the third side and all the other angles, but I'm not quite sure how to find the actual area of this triangle. 3) Find the sixth roots of -64. I learned how to do this, but the entire theory doesn't really make sense. Thank you. Date: 07/19/98 at 20:29:11 From: Doctor Jaffee Subject: Re: Trig Hi Susie, You have asked 3 very good and diverse questions so this is going to take some time, but I think I can help you, and I hope provide you with a better understanding of the processes involved. First of all let's try the airplane problem. Vectors can be used to represent the directions and speeds of the three forces involved in this problem, namely the airplane when there is no wind, the wind, and the airplane with the wind. Draw a triangle using three vector arrows. Make vector AB horizontal and label its length 20.4. The arrowhead should be to the left (west) since the wind is coming out of the east. Make vector AC vertical and long enough so that the length of the hypotenuse will be 241. The hypotenuse, vector BC, represents the speed and direction of the airplane if there were no wind. You should now be able to see that the direction that the plane must fly in order to end up on a due north course is determined by the angle ABC. You should be able to determine the measure of that angle easily by using the fact that cos B = AB/BC = 20.4/241 and you can use the Pythagorean Theorem to determine the length of vector AC, the speed of the airplane. Now on to the second problem. Anytime you know two sides of a triangle and the included angle you can always find the area by multiplying the lengths of the two sides by the sine of the angle, then dividing the product by 2. Here's how I know. Draw the triangle you described in the problem and draw the altitude BD. The point D is outside the triangle on the line AC. Now, the sine of angle BAD and the sine of angle BAC are the same. (I'll leave that up to you to justify, but in the meantime take my word for it). So, sin BAD = BD/c, where BD is the height of the triangle. Let's rename the height with just the letter h. Therefore sin BAD = h/c and sin BAC = h/c. Now, if you multiply both sides by c, you get c*sin BAC = h. Normally, the way you find the area of a triangle is base * height/2. But if we replace "base" with b and height with c*sin BAC, we get the result I predicted. Finally, the sixth roots of -64. DeMoivre's theorem is the easiest way I know to solve this one, and you have to put the number -64 into complex polar form to make it work. Make a grid in which the main horizontal axis is a real number line, positives to the right, negatives to the left, and 0 where it intersects the vertical axis. The vertical axis has imaginary numbers: i,2i,3i, etc above 0, and -i,-2i,-3i etc. below 0. The number -64 would then be located 64 units to the left of 0. Now imagine a vector with initial point at 0 and terminal point at -64. The length of this vector is 64 and the angle that the vector makes with the positive real axis is 180 degrees. Now, a number in complex polar form looks like r(cos T + isinT), where r is the length of the vector from the origin to the number and T is the angle between the vector and the positive real axis. Therefore, -64 in complex polar form is 64(cos 180 + isin 180). DeMoivre's Theorem states that if you want to take the sixth root of a number in complex polar form you must take the sixth root of r, then divide T by 6. Since the sixth root of 64 is 2 our result is 2(cos 30 + isin 30), which you can convert into rectangular form because you know that cos 30 = SQRT(3)/2 and sin 30 = 1/2. That's one of our six answers. To get our second answer we use the fact that if you go around a circle 180 degree or 180 + 360 = 540 degrees you end up in the same location. Let's rename our number 64(cos 540 + isin 540). Apply DeMoivre's Theorem again and get 2(cos 90 + isin 90). If we add another 360 degrees to our original value of T we can go through the same procedure and get a third solution, and so on. You'll notice that we can't do this forever because the seventh solution is actually identical to the first solution. I hope that I have helped you understand these problems a little better. If not, or if you have other questions write back. Also, take a look at the archives, where there are a lot of questions similar to the ones you asked. - Doctor Jaffee, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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