1 cos. cos cos cos cos MAT 126H Solutions Take-Home Exam 4. Problem 1

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1 MAT 16H Solutions Tke-Home Exm 4 Problem 1 ) & b) Using the hlf-ngle formul for cosine, we get: 1 cos cos cos 8 4 nd 1 8 cos cos 16 4 c) Using the hlf-ngle formul for tngent, we get: cot ( 3π 1 ) = cot ( π 1 ) = 1 tn ( π 1 π + cos ( = [1 6 ) ) sin ( π 6 ) ] = [ ] = 3 d) Let α = sin 1 ( 3 ), then sinα = 3, nd cosα = 4 (since α QI). We then hve: cos ( 1 sin 1 ( 3 5 )) = cos ( α ) = 1+cosα = = 9 1 e) If z i 1, then z is therefore given by x,y 1,1 r,, 3 4. The trigonometric (or polr) form of z i 1 cos 3 4 isin 3 4.

2 Applying De Moivre s Theorem, we then get z 1 1 cos isin cos 15 isin 15 5 i 1 3i. Problem ) [ASA cse] Since B , the Lw of Sines yields sin 45 sin 95 4sin 45 c.84. c 4 sin 95 b) [SAS cse] By the Lw of Cosines we hve c 3 (3)() cos5 c 131cos5.3. Therefore, the perimeter of ABC is b c c) [SSA cse] Here the Lw of Sines yields sin 7 sin A 5sin 7 sin A 1.57 but this is impossible since sin 1 for ny ngle! Therefore, we conclude tht no tringle ABC cn stisfy these conditions. Also, the fct tht exist. 3 h 5sin implies tht the tringle ABC does not

3 d) [SSA cse] This is the so-clled mbiguous cse with two possible tringles ABC (one with ngle A obtuse, the other with ngle A cute). The two possibilities re given below. 1) If ngle A is obtuse, then the Lw of Sines yields sin sin C 6sin sin C C 3.9. Therefore A ) If ngle A is cute, then C nd so we hve A e) [SSS cse] First, note tht ABC is isosceles since = b = 3. Thus, the mesure of A + B is equl to the mesure of A (or B). Applying the Lw of Cosines yields Therefore, A. cos A 7.53 ()(3) 3 A B A (7.53 ) Problem 3 Suppose the distnce between the bot nd the cliff is d. We then hve tn d 14.45'. d tn 5 Sketch right tringle with bse hypotenuse. d, height 1, nd the ngle mesuring 5 between d nd the

4 Problem 4 First, compute the re of the tringulr lot by pplying Heron s formul. Since the semiperimeter is s 75 m, we get: m A Multiplying this re by 35 (the price per squre meter) yields finl price for this tringulr lot of $3,451. (rounded to the nerest dollr). Problem 5 Let x be the direct (stright) distnce from Ft Myers to Orlndo. By the Lw of Cosines, we then hve x cos 13 o x 3 miles. Sketch tringle with sides of length 15, 1 nd x, with ech vertex representing city. The ngle opposite the side with length x hs mesure of 135 degrees bsed on the clockwise turn towrds Orlndo from Srsot. Problem 6 ) Let h be the height of the tringle perpendiculr the side of length. Using the right tringle with legs h nd nd hypotenuse b, we get cos(b) = /b. b) By the Lw of Cosines we hve: cosb b b b cos B b.

5 c) By the Lw of Sines nd the difference formul for sine, we get: sinb b sin A sin 18o B sin18ocos B cos18 o sin B sin B sinbcos B 1 b cosb cos B b. Problem 7 Applying Heron s formul to this isosceles tringle, we get b b b b b A b b b 4b. 4 4 Note: You cn lso find this re by pplying the Pythgoren Theorem to the right tringle with leg nd hypotenuse b. Problem 8 ) Let d be the distnce between the two towers. We then hve d = 1776 tn(34 ) 633. b) Let h be the height of the office building. Looking t the right tringle with bse d, height 1776 h, nd ngle opposite this height mesuring, we then hve tn( ) = 1776 h d 1776 h = 1776tn( ) tn(34 ) h = 1776 (1 tn( ) tn(34 ) ) 818

6 Problem 9 Let x be the length AD of the new rmp. Then the problem is illustrted by the digrm bove. Looking t the right tringle ABC, we see tht ngle ˆ ABC mesures Looking t tringle ACD, we see tht ngle mesures ˆ DAC mesures nd so ngle ADC ˆ By the Lw of Sines, we then hve sin 7 sin1 1sin 7 x 9.7 x 1 sin1 We conclude tht the new rmp mesures, pproximtely, 9.7 feet. Note: You cn lso solve this problem by using the right tringles ACD nd ADB.

7 Problem 1 From the illustrtion bove, we see tht the polygon ABCDE (n pproximte covering of the lke) cn be prtitioned into the 3 tringles ABE, BDE, nd BCD, whose respective res we shll denote by A 1, A, nd A 3. We then need to compute ech of these 3 res: 1 A1 (1)(15)sin m 1 A3 (7)(5)sin1 173 m To compute A, we first find the lengths BE nd BD using the Lw of Cosines: BE 1 15 (1)(15) cos m BD 7 5 (7)(5) cos1 9.8 m Next we use Heron s formul: s (the semi-perimeter of BDE) So A s( s 5)( s 97.75)( s 9.8) 87m. We conclude tht the totl re of the lke is given, pproximtely, by A1 A A3 8798m. Note: To void rounding errors, keep exct forms s long s possible. When pproximtion must be done on intermedite results, use t lest or 3 more significnt figures thn wht is sked in the question. For exmple, in the problem bove round distnces BE nd BD to the nerest hundredth of meter, nd round your tringulr res to the nerest hundredth of squre meter.

8 Problem 11 ) The modulus of z is given by r Since x rcos nd y rsin, we hve cos nd Thus, 11 nd the trigonometric form of z is given by 6 z 3 cos 11 isin cos 6 isin 6. sin b) Applying De Moivre s Theorem, we get 6 cos 6 6 z 6 3 isin As result, 6 z 178. Bonus Problem The three complex cube roots of z re given s follows: 3 z 3 cos 11 isin 11 ( k ) z 3 cos 3 isin 3 ( k 1) z 3 cos 35 isin 35 ( k ) 18 18