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Helpppp!!! Me With This Trig Problem Step By Step Please

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lacticacid | 21:02 Sat 02nd May 2015 | Jobs & Education
6 Answers
In triangle ABC side a is across from angle A, side b is across from angle B and side c is across from angle C.

If A = 14 degrees, B = 26 degrees and c = 4, find a, b and C.

Answers should be rounded to 3 decimal places.

Show at least the initial equations you set-up to find sides a and b.
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Angle C is easy. The internal angles of all triangles add up to 180 degrees. Therefore if A+B+C=180, C=180-A-B. C=180-14-26. C=140 degrees.
Given two angles in a triangle you can always find the third through the equation A + B + C = 180. Having done so, two applications of the sine rule in the form

a/(sin A) = c/ (sin C)

to find a

b/(sin B) = c/ (sin C)

to find b

will give the remaining sides. Finally, just plug in the numbers into your calculator and make sure it is set to degree mode.
If you haven't figured out from all the other trig questions you've posted that if it's a right-angled triangle you use Sin, Cos or Tan and if it's not like this one you use the Sine Rule or Cosine Rule then I don't think you should be even trying the questions as you're clearly not ready.
jim360 - I think from the OPs other 'homework' questions on here, you may have left him/her way, way, way behind with your excellent answer :-))
Very possibly, Captain. At the moment I can't think of much else to add. At this level of mathematics it's very algorithmic really.

For a triangle, ask yourself what information do you know, and follow the "flowchart":

i) all three sides a,b,c? Cosine rule in form

Cos A = (b^2 + c^2 - a^2)/ (2 b c )

ii) any two angles A, B and one side?

iia) if the given side is a or b use sine rule straight away to find other sides:

a/(sin A) = b/(sin B) = c/ (sin C)

iib) if the given side is c (ie two angles at either end of the given side) then use sum rule for angles 180 - A - B = C and then use sine rule (as given in this question).

iii) two sides a, b and the angle between them C? cosine rule in the form

c^2 = a^2 + b^2 - (2 a b) Cos C

iiib) two sides in a right-angled triangle? Special case of above, Pythagoras Theorem

c^2 = a^2 + b^2

iv) two sides and one of the angles not between them? Sorry, you're buggered (two solutions possible).

v) all three angles? Sorry, you're buggered. (Similar triangles).

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