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"
Also Give checks.
I know the use of logarithms is unnecessary but i have to show my solution by using logarithms and that's the only reason I'm asking for help. Logarithms confuse me. Please help me.
If you are not already familiar with the law of tangents, here's an article that can help you:
http://en.wikipedia.org/wiki/Law_of_tangents
We will be using the same notations in the following solution.
In the given case, sides a,b are known, and the included angle C. So the sum of angles
A+B=180-C = 180 - 32-28-30 = 147-31-30 = 1967π/2400 radians
a+b=21.46+46.28=67.74
a-b=21.46-46.28=-24.82
By the law of tangents,
tan(A-B)/tan(A+B)=(a-b)/(a+b)
or
tan(A-B)=tan(A+B)(a-b)/(a+b)
=0.2331984
In log (to base 10), cannot calculate the values of negative numbers, so we will keep track of the sign ourselves:
log(-tan(A+B)) = -0.1962309
log(-(a-b)) = 1.394801777162711
log(a+b) = 1.830845192308612
log(tan(A+B)*(a-b)/(a+b) = -0.1962309+1.3948018-1.8308452
=-0.6322743
Antilog(-0.6322743)=0.2331985 as before.
Convert 0.2331985 to degrees,
A-B=13-21-41, and
A+B=147-31-30
A = (160-53-11)/2=80-26-36
B = (134-09-49)/2=67-04-55
side c can be found by the cosine rule or the sine rule.
Check my work.
Anon & Drws, thank you for pointing out there is a problem with this solution.
The tangent rule formula that I used was not correct. The formula used for the tangent rule is simply:
(a-b)/(a+b) = tan((A-B)/2)/tan((A+B)/2)
Using the previous values,
a=21.46
b=46.28
C=32-28-30=32.475
We have
A+B=180-32.475=147.525
a+b=67.74
a-b=-24.82
Now apply the tangent rule:
tan((A+B)/2)=3.433633
tan((A-B)/2)=tan((A+B)/2)*(a-b)/(a+b)
= -1.2580863
(A-B)/2 = atan(1.2580863)= -51.520285
A-B = -103.04057
A=(147.525-103.04057)/2=22.242215°
B=(147.525+103.04057)/2=125.282785°
c can be found by the sine rule or the cosine rule, since 5 of the six unknowns have been calculated. I get c=30.44083 using the cosine rule and it checks with the sine rule.
I believe you can now handle the logarithm part. If you need further help, just post.
I apologize again for the unforgivable mistake in applying the tangent rule formula.
To solve the triangle ABC using logarithms and the law of tangents, you will need to use the following formula:
tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B)),
where A and B are the angles opposite to the sides a and b, respectively.
Here are the steps to solve the triangle:
Step 1: Convert the angle C from degrees, minutes, and seconds to decimal degrees.
C = 32 + (28/60) + (30/3600)
= 32.475 degrees
Step 2: Use the law of tangents to find angle A:
tan(A) = (b * tan(C)) / (a - b * tan(C))
tan(A) = (46.28 * tan(32.475)) / (21.46 - 46.28 * tan(32.475))
Now, find the value of A by taking the inverse tangent (or arctan) of both sides of the equation and using logarithms:
A = arctan[(46.28 * tan(32.475)) / (21.46 - 46.28 * tan(32.475))]
Use logarithms to compute the arctan:
A = log[(46.28 * tan(32.475)) / (21.46 - 46.28 * tan(32.475))]
Step 3: Use the law of tangents to find angle B:
tan(B) = (a * tan(C)) / (b - a * tan(C))
tan(B) = (21.46 * tan(32.475)) / (46.28 - 21.46 * tan(32.475))
Now, find the value of B by taking the inverse tangent of both sides of the equation and using logarithms:
B = arctan[(21.46 * tan(32.475)) / (46.28 - 21.46 * tan(32.475))]
Use logarithms to compute the arctan:
B = log[(21.46 * tan(32.475)) / (46.28 - 21.46 * tan(32.475))]
Step 4: Calculate the third angle, C:
C = 180 - A - B
Now you have the values of angles A, B, and C. To check your solution, you can calculate the side lengths using the law of sines or cosines and compare them to the given side lengths a, b. If they match, it means your solution is correct.
Please note that logarithms are not necessary for solving this triangle; they are used here to fulfill your requirement.