please help /trigonometry?

Question

plz show work so i can understand. thanks


Note that the point value of each question is given in parentheses and give checks for the following question.

9. (10) Use logarithms and the law of tangents to solve the triangle ABC, given that a= 21.46 ft, b= 46.28 ft, and c= 32º 28’ 30”

Answer 1

(Check Wikipedia for a full explanation)

The law of tangent is used when you are doing a calculation involving two angles and the two sides opposite those angles.

In triangle ABC, side a is opposite angle A, side b opposite angle B and side c opposite angle C

(common usage is lower case letters for sides and upper case letters for angles -- Wikipedia uses Greek letters for angles)

Law of tangents:

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

Knowing any three of these elements, you can find the fourth.

However, in this case, it appears that you are not given the proper angle. Instead, you are given the angle between the two sides. Such a case would be better solved by the law of cosines.

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Back to law of tangents, Wikipedia shows a worked out example given two sides and one angle. Once you find the second angle, you can quickly find the third angle, knowing that the sum of the three angles must be 180˚.

Once you have all three angles and two of the three sides, then you can use any of the three laws to find the third side.

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As far as using logarithms: if two number are to be multiplied, you find their logarithms, add the logarithms, then find the antilog. For divisions, subtract the logs.

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

can be rewritten:



{(a-b) [tan(A+B)/2]} / {(a+b) [tan((A-B)/2)]} = 1

Then, using logarithms:

log( {(a-b) [tan(A+B)/2]} / {(a+b) [tan((A-B)/2)]} ) = log (1)

a division = subtract the logs; also, log(1) = 0 in any base:
log{(a-b) [tan(A+B)/2]} - log{(a+b) [tan((A-B)/2)]} = 0

each multiplication = add the logs:
{log(a-b) + log[tan(A+B)/2]} - {log(a+b) + log[tan((A-B)/2)]} = 0

removing the braces { }:
log(a-b) + log[tan(A+B)/2] - log(a+b) - log[tan((A-B)/2)] = 0

However, logarithms are not very efficient in solving the law of tangents.

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ahh! A+B = 180 - C.

Then in our equations, for example,
log(a-b) + log[tan(A+B)/2] - log(a+b) - log[tan((A-B)/2)] = 0

we know (a-b), (a+b) and (A+B). therefore we can find three of the four terms.

The fourth term log[tan((A-B)/2)] is the value needed to make the equation come out to 0.

Once you have found log[tan((A-B)/2)]
use antilog to find tan((A-B)/2)
use arctan to find (A-B)/2
multiply by 2 to find A-B

Then you'll have two equations:
A + B = 180 -C
A - B = whatever you just found (call it x for now)

Add the two equations, the B cancels out:

2A = 180 - C + x

A = (180 -C + x)/2