Expert: Sherman D. - 12/1/2006

Question

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The text above is a follow-up to ...

-----Question-----
1). Simplify the following expressions so that they involve a function of only one triangle:

A.sin 80°-sin 10°/sin 80° +sin 10°

B. sin 130°+sin 20°/cos 130°+cos 20°

2). Use logarithms and the law of tangents to solve the triangle ABC, given that a=21.46 ft, b=46.28 ft, and C=32°28'30".

3). Solve the the triangle for which the given parts are a=27, b=21, and c=24.

Thank you.
-----Answer-----
A.)
(sin(80) - sin(10))/(sin(80) + sin(10)) = (

sin(x) - sin(y) = 2sin((x - y)/2)cos((x + y)/2)
sin(x) + sin(y) = 2sin((x + y)/2)cos((x - y)/2)

sin(80) - sin(10) = 2sin((80 - 10)/2)cos((80 + 10)/2)
sin(80) - sin(10) = 2sin(35)cos(45)
sin(80) - sin(10) = 2sin(35)(sqrt(2)/2)
sin(80) - sin(10) = sin(35)sqrt(2)

sin(80) + sin(10) = 2sin((80 + 10)/2)cos((80 - 10)/2)
sin(80) + sin(10) = 2sin(45)cos(35)
sin(80) + sin(10) = 2(sqrt(2)/2)cos(35)
sin(80) + sin(10) = cos(35)sqrt(2)

(sin(35)sqrt(2))/(cos(35)sqrt(2))
(sin(35))/(cos(35))
tan(35)

(sin(80) - sin(10))/(sin(80) + sin(10)) = tan(35)

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B.)
(sin(130) + sin(20))/(cos(130) + cos(20))

sin(130) + sin(20) = 2sin((130 + 20)/2)cos((130 - 20)/2)
sin(130) + sin(20) = 2sin(75)cos(55)

cos(130) + cos(20)) = 2cos((130 + 20)/2)cos((130 - 20)/2)
cos(130) + cos(20) = 2cos(75)cos(55)

(sin(130) + sin(20))/(cos(130) + cos(20)) =
(2sin(75)cos(55))/(2cos(75)cos(55)) = (sin(75))/(cos(75)) = tan(75)

(sin(130) + sin(20))/(cos(130) + cos(20)) = tan(75)

I had to use google to search for sin(x) + sin(y) and cos(x) + cos(y), if you try looking up sin(x) - sin(y), you won't be able to get anything. It just like sin(x) + sin(y), only the signs are changed.

It has been so long since i did problems like this and i think you for allowing me to redo problems like them. If you notice, i like doing trig. That is why i asked below for your help on the logarithmic trigonometry problems.

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2.)
On this one, i haven't done logarithms and trig, but if you can tell me what log(tan), log(sin), log(cos), and their inverses are, i would be very greatful, because i have been trying to remember how to work them. I believe i use to work them when i was in school, but i have forgotten. The reason i am asking help from you is in case i have any other questions like this one. I have tried looking it up and i couldn't find anything that was really all that helpful.

For ex:

log(tan(A)) = log(sin(L)) + log(tan(HA)) = log(sin(L) * tan(HA))

log(cos(A)) = (log(sin(D)) - log(cos(L)) = (cos(A) - sin(D))/(cos(L))

but i am not certain what log(sin(A)) is or what L, HA, or D is suppose to be. I forgot how to do it, but if you can help. I already know the law of tangents, just now how to work the log of it.

This is the book that im studying from it has the questions at the bottom. http://lessons.educationdirect.com/pdf/6232d.pdf im not to sure about log just started trigonometry 1 week ago but heres some more info on that http://www.sundials.co.uk/tbcal.htm
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3.)
Law of Cosines

27^2 = 21^2 + 24^2 - 2(21 * 24)cos(A)
729 = 441 + 576 - 1008cos(A)
729 = 1017 - 1008cos(A)
-288 = -1008cos(A)
cos(A) = (288/1008)
A = 73°23'54.42"

441 = 729 + 576 - 2(27 * 24)cos(B)
441 = 1305 - 1296cos(B)
-864 = -1296cos(B)
cos(B) = (864/1296)
cos(B) = (2/3)
B = 48°11'22.87"

576 = 441 + 729 - 2(27 * 21)cos(C)
576 = 1170 - 1134cos(C)
-594 = -1134cos(C)
cos(C) = (594/1134)
C = 58°24'42.71

A = 73°23'54.42"
B = 48°11'22.87"
C = 58°24'42.71

Answer
www.sundials.co.uk/tbcal.htm is where i got my info, i just didn't understand what was what.

On A and B, you will still get the same answer. Is the way i did it ok, or are you suppose to have it in the form that the pdf website showed me.

2). Use logarithms and the law of tangents to solve the triangle ABC, given that a=21.46 ft, b=46.28 ft, and C=32°28'30".

Thanks for the pdf site, now i can help you

a = 21.46
b = 46.28
C = 32°28'30"

Using Law of Tangents

(b - a)/(b + a) = (tan((B - A)/2))/(tan((B + A)/2))

(46.28 - 21.46)/(46.28 + 21.46) = (tan((B - A)/2))/(tan((B + A)/2))

(24.82/67.74) = (tan((tan((B - A)/2))/(tan((B + A)/2))
(1241/3387) = (tan((B - A)/2))/(tan((B + A)/2))

Since we have C, we can find B + A

(B + A) + C = 180
(B + A) + 32°28'30" = 180
B + A = 147°31'30"

Using log(tan((B - A)/2))

log((tan((B - A)/2)) = log(b - a) - log(b + a) + log(tan((B + A)/2)))

log((tan(B - A)/2)) = log(46.28 - 21.46) - log(46.28 + 21.46) + log(tan(73°45'45"))

log((tan(B - A)/2)) = log(24.82) - log(67.74) + log(tan(73°45'45"))

log((tan(B - A)/2)) = .099710433
tan((B - A)/2) = 10^.099710433
(B - A)/2 = tan^-1(10^.099710433)
B - A = 2tan^-1(10^.099710433) or 103°2°26.05

Now using this, you get

B - A = 103°2'26.05"
B + A = 73°45'45"

2B = 176°48'11"
B = 88°24'5.53"

88°24'5.53 + A = 73°45'45
A = -14°38°20.53

When you figure out where i went wrong, let me know. Since side "a" is smaller than side "b", it's confusing me.

I used both the log(tan((B - A)/2)) and the law of tangents and i still get B - A = 103°2'26.05"

Sorry i couldn't help. If you go to answers.yahoo.com and give them the pdf site, maybe they can help you more.

Thanks for that website, it showed me info on stuff i don't have yet.