Solutions to Assignment #07 MATH radians = = 7 (180 ) = 252 : 5

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1 Solutions to Assignment #0 MATH 0 Precalculus Section. (I) Comlete Exercises #b & #0b on. 0. (#b) We robabl need to convert this to degrees. The usual wa of writing out the conversion is to alwas multil b a fraction equal to one. radians 0 radians 0 : This is located in Quadrant III because 0 < < 0. Since radian measure does NOT require us to write the radians tag on the angle (radian measure is technicall unitless!), we can also shortcut the rocess if the radian measure exression has in it: (0 ) : If it doesn t have in it, then we must still multil b (0) : (#0b) Which quadrant? (0 ) 0 : We need to add a whole revolution to this angle (or an convenient multil of 0 ). Thus, we have the coterminal (this will ut us at the same oint on the unit circle) as : This is located in Quadrant II. Another wa to look at similar roblems is to determine the number of revolutions before we convert to degrees. radians. Since we know that is one revolution, we see that if we round u can add one revolution easier in radians. + (II) Comlete Exercises #ab, #ab & #ab on.. 0 : (#a) The comlement of is (0 ) : Thus, the comlement of is : The sulement of is (0 ) : Thus, the comlement of is : (#b) The comlement of is (0 ) : The sulement of is (0 ) : to ; we

2 (#a) Multil b (0 ) : (#b) Same. 0 : 0 0 : (#a) Since these have in them, we can substitute 0 each time. (#b) Same. (0 ) 0 0 : (0 ) 0 : Note that since this is greater than 0 ; this shares a coterminal with (0 0 ) : (III) Comlete Exercises # & #0 on. -. (#) Since Lnchburg is due north of Mrtle Beach, the two cities must lie on a great circle of the Earth. If we assume the Earth is a erfect shere (it isn t...), then an lane which asses through the center of the Earth will create a great circle intersection with the shere. In this case, the great circle includes the North and South Poles at the to and bottom of the great circle, resectivel. The radius of the great circle must be the same radius for the shere, km. So from the side, the Earth looks like this (the scale is one unit 000 km): 0 0 x In this context, a ositive angle in standard osition reresents north latitude. I ve sketched in a radius which intersects the Earth s great circle in the northern hemishere at 0 N latitude. It we have an acute angle below the ositive x-axis, then that negative angle reresents south latitude. Thus, when one cit is due north (or south) of another, the must both lie on the same great circle. Now we can al our radian measure formulas. Since the distance measured on the Earth s surface is reall a iece of (great circle) arc, we have (in radians) s r : Thus, we have 00 km km : 0:0 radians.

3 [Note that, in general, whenever Earth measurements are given in radians, we usuall use SIX decimal laces of accurate (a microradian).] We convert this to degrees and this will be the di erence between their latitudes. 0 : 0:0 : : Also, here s a lesson on how to change degree measure with decimal laces to degree measure with minutes and seconds of angle arc. First, take our exression and stri o the whole number of degrees. 0 0:0 : 0:0 degrees. Since there are 0 minutes in a degree, multil b 0. 0 (0:0) : :0 minutes. Stri o the minutes, and the multil b 0 to obtain the remaining number of seconds of angle arc. 0 (0:0) : :0 seconds. Thus, our original number of degrees latitude (di erence) is 0 : 00 : (#0) We assume that a.-inch circular ower saw blade has a.-inch DIAMETER (not radius). Section. Thus, its radius is r inches. (a) At 00 rm, the angular seed (omega) in RADIANS PER SECOND is 00 rev min radians rev min 0 sec 0 radians er second!: (b) We convert angular seed to linear seed b multiling b the radius. The units of the radius take the lace of the radian measure since the radian measure does not reall have an units attached to it. 00 rev min radians rev foot inches inch feet er minute. This is arox. 0 feet er minute which is about mh! (IV) Comlete Exercise # on.. We have t : This is a Quadrant IV angle. Since the denominator is, the reference angle must be 0 : This means that the cosine and sine values (osition on the unit circle) must be! (cos () ; sin ()) ; :

4 Since the angle is in Quadrant IV, the cosine (x-coordinate) must be ositive and the sine (-coordinate) must be negative. This gives us cos () ) sec () cos () sin () ) csc () sin () : Ticall, we take the ratio of sine and cosine to obtain the tangent value. tan () sin () cos () (V) Comlete Exercise # & # on.. (#) We see that is more than since ) cot () : tan () ( ) : B eriodicit; we obtain a smaller coterminal b subtracting an convenient multile of : Thus, we have cos cos + cos cos (0 ) : This is a Quadrant II angle. Since its denominator is (when the fraction is in lowest terms), the reference angle must be 0 and its cosine and sine values must be (cos () ; sin ())! ; : Since lands us in Quadrant II, we must have a ositive cosine. Thus, we have cos + : (#) Same. We have : So if we subtract ; we have sin sin + sin sin (0 ) : Thus, we land in Quadrant III. The denominator is so the reference angle is 0 and the cosine and sine values are! (cos () ; sin ()) ; : In Quadrant III, both the cosine and the sine are negative. sin sin : Thus, we have

5 (VI) Comlete Exercise #0ab on.. (a) We have cos (t) : We do NOT need to know the exact angle here. We know that the angle is either in Quadrant II or Quadrant III. Recall that the cosine function must be even. cos ( t) cos (t) This is because if we go in the ositive (counterclockwise) direction for an angle in standard osition, we will end u at the same x-coordinate if we go the same angle distance in the negative (clockwise) direction. Thus, we have cos ( t) : (b) The secant is alwas the recirocal of the cosine. Section. sec (t) cos (t) : (VII) Comlete Exercise # on. 0. We assume that these angles are in Quadrant I. sec () h adj : Thus, we can draw a right triangle in standard osition with hotenuse c and side adjacent a : Thus, b Pthagoras, the side oosite must be a + b c ) + b ) b ) b : We know that this must be ositive in Quadrant I. This gives us cos () sec () sin () o h ) csc () sin () tan () o adj ) cot () (VIII) Comlete Exercises # & # on. 0. (#) Start on the left side. Distribute. We substitute cos () + sin () : tan () : ( + sin ()) ( sin ()) sin () + sin () sin () sin () cos () + sin () sin () cos () :X

6 (#) The common denominator is cos () sin () : sin () cos () + cos () sin () sin () sin () cos () cos () + cos () sin () cos () sin () sin () + cos () cos () sin () cos () sin () : The trick is to searate this into a roduct! cos () sin () cos () sin () (IX) Comlete Exercises # & #0 on. 0. (#) Two arts. (a) If ou like doing this right triangles, then we have cot () adj o : sec () csc () :X Thus, create a right triangle in standard osition with adjacent side a and oosite side (vertical leg) b : This gives us the hotenuse c + + ) c : The cosine and snie values would be cos () adj h The angle must be : Alternate solution: Use the identit o ; sin () h : + cot () csc () ) +! + csc () csc () r ) sin () ) : (b) If sec () ; then cos () ) : (#0) In the right triangle, we are given 0 in standard osition and the hotenuse is. The missing side is the side oosite. sin (0 )! ) sin (0 ) :

7 (X) Comlete Exercise # on. 0. Indirect measurement. Suose we dro a erendicular from the mountain eak down to car-level. Let that distance be ; since that is our main unknown and it is a vertical segment. We note that in real life, it would imossible to make a direct measurement unless we could drill directl down into the mountain! It turns out that we need the distance between the angle vertex and the erendicular. Let that distance be x since it is a horizontal segment. This gives us two right triangles. In the case of the : angle, we have In the case of the angle, we have tan (: ) x + : tan ( ) x : So it turns out that it s reall eas to nd x b dividing the equations b each other. If we cross-multil, we obtain tan (: ) tan ( ) tan (: ) (x + ) tan ( ) x tan (: ) x + tan (: ) tan ( ) x x+ x x x + tan (: ) tan ( ) x tan (: ) x (tan ( ) tan (: )) x x tan (: ) tan ( ) tan (: ) : : miles. From the second equation, we have This is arox. feet. tan ( ) x ) tan (: ) tan ( ) tan ( ) tan (: ) : : miles.