6232C - Answers to PF Exam number : 6232C PLANE...

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Unformatted text preview: Answers to PF Exam number : 6232C PLANE TRIGONOMETRY (6232C) 1) Find the Sine, Cosine, Tangent and cotangent of: a) Sin 122o = 0.848; cos 122o = ­0.530; tan 122o = ­1.600; cot 122o = ­0.625 b) Note: 315o 12' = 315.2o Sin 315o 12' =­0.705; cos 315o 12' = 0.710; tan 315o 12' = ­0.993; cot 315o 12' = ­1.007 c) 275o 13' 37" = 275.2269444o Sin 275o 13' 37" = ­0.996; cos 275o 13' 37" =0.091; tan 275o 13' 37" = ­10.931; cot 275o 13' 37" = ­0.915; d) 193o 41' 51" = 193.6975o Sin 193o 41' 51" =­0.237 cos 193o 41' 51" = ­0.972; tan 193o 41' 51" = 0.244; cot 193o 41' 51" = 4.103; 2) Using exact values find the numerical values of a) Sin 30o cos 240o +sin 210o sin 300o = (1/2)*(­1/2) + (­1/2) (­√3/2) = ­ (1/4) + (√3/4) = (√3­1)/4 b) Tan 225o + tan (­45o) = (1) + (­1) = 0 3) Express each of the following as same function of a positive acute angle: Note: sin(a) = sin(180o ­ a); cos(a) = ­cos(180o ­ a); tan(a) = ­tan(180o ­ a); sin(­a) = ­sin(a); cos(­a) = cos(a); tan(­a) = ­tan(a) a) Cos 112o 33' = ­cos 67o 27' b) Tan 310o = tan (­50o) = ­tan 50o c) Cot 138o 15' 10" = ­cot 41o44'50" d) Sin (­140o) = ­sin 140o = ­sin 40o 4) Express each as a function of θ a) sin (270o + θ ) sin (A+B) = sin (A) cos (B) + sin (B) cos (A) Using the formula, we can find that Sin (270°+θ) =sin270°cosθ+sinθcos270° but cos 270° = 0 and sin270° = ­1 So sin (270°+θ) = ­cos θ b) Cos ( π + θ ) Using formula Cos(A+B)= cos(A) cos(B)­sin(A) sin(B) = cos( π )cos θ ­ sin( π )sin θ = ­1cos θ ­ 0 = ­cos θ Cos ( π + θ ) =­cos θ c) Tan (810o + θ ) using formula Tan (A+B) = tan A + tan B tan 810 + tan θ − 0.5871 + tan θ = = 1 − tan( A) tan( B ) 1 − tan(810)(tan θ ) 1 + 0.5871 tan θ d) Sin ( θ ­180o) = sin θ cos 180o – cos θ sin180o = ­sin θ 5) Find the value of: a) Log sin 62o 22' 33" = ­0.053 b) Log cot 28 o 13' 17" = 0.270 c) Log cos 125 o 15' 23" = ­0.577 No answer, because it is a negative number and you can't take log of a negative number. d) Log tan 78o 45' 50" = 0.702 6) Find the acute angle A, to the nearest second when: There are no solutions to (a) or (b) as I understand the problem ­ as you can see in the work below, finding A is out of the domain range because cos of any angle could never equal 109.12575 ­ a number much much bigger than 1. Am I misreading the problem? I believe this is log10(cosA) = 9.12575. Is this correct??? Please email me back and we will see if we can figure this out. a) Log cos A=9.12575 109.12575 = cosA A = cos­1 109.12575 A = b) Log sin A=9.91655 109.91655 = sinA A = sin­1 109.1655 A = c)log cot A=0.11975 100.11975 = cotA A = cot­1 100.11975 A = 0.18939 = 0o 1' 8" d)log tan A=0.06323 100.06323 = tan A A = tan­1 100.06323 A = 49.156259 = 49o 9' 22" 7) Simplify sin (90 o + x) sin (180 o + x) cos (90 o + x) cos (180 o ­ x) =[sin(90 o)cos(x) + cos(90 o)sin(x)]* [sin(180 o)cos(x) + cos(180 o)sin(x)]*[cos(90 o)cos(x) ­ sin(90 o )sin(x)]*[cos(180 o)cos(x) + sin(180o)sin(x)] = [cos(x) + 0]*[0 ­ sin(x)]*[0 ­ sin(x)]*[­cos(x) + 0] = ­sin2(x) cos2(x) 8) Solve the right triangle by logarithms: A=28 o 30', b=18.3 B = 61.5o or 61o 30' C = 900 Side a = 9.936 Side c = 20.823 Solve the following oblique triangles 9) a =31,b=15, c=17. Using law of cosines: 2 2 2 c =a +b ­2ab cos(C) 172 = 312 + 152 ­ 2(31) (15)cos(C) 1054cosB = 1025 cosB = 0.972 B = 13.47o 312 = 152 + 172 ­ 2(15)(17)cosA ­510cosA = 447 cosA = ­0.876 A = 151.22o C = 15.31o 10) B = 115o 30' , C = 20o 29' ,a=23.47. A = 44o 1' Sin (115o 30') / b = sin (44o 1) / 23.47 b = sin(115o 30')*23.47 / sin(44o 1) b = 30.486 Sin (20o 29') / c = sin (44o 1) / 23.47 c = sin(20o 29')*23.47 / sin(44o 1) c = 11.819 11) a = 134.2,b = 84.54, B = 52o 9' 11" Sin (52o 9' 11") / 84.54 = sin (A) / 134.2 Sin (A) = 134.2* Sin (52o 9' 11”) / 84.54 A = Does not exist This triangle is not possible. 12) a = 627.7, b = 412.2, A = 66o 47' Sin66o 47' / 627.7 = sin(B )/ 412.2 Sin(B) = 412.2* sin66o 47' / 627.7 = 0.6035 B = 37.12o = 37o7'17" C = 76.095o = 76o 5' 43" Sin 66o 47' / 627.7 = sin37o7'17" / b b = 627.7 * sin37o7'17" / sin66o 47' b = 412.19 ...
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This note was uploaded on 03/21/2011 for the course BUS 123 taught by Professor Professorjames during the Spring '11 term at Central Pennsylvania.

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