You can check your value of c by using the Law of Cosines, with the original calyues of a, b and C.
c^2 = a^2 + b^2 -2ab cos C
c^2 = 460.32 +2141.84 -1986.34*0.8436
c = 30.44
That is not what you came up with. There appears to be something wrong with your initial angles. The law of sines does not agree with your angles A and B. If A > B, than you should have a > b
Perhaps Mathmate can explain the origin of the problem.
I want to start by saying thank you . You have no idea how much u have helped me understand logarithms, even better then the books i have (it poorly explains the subject of trigonometry let alone logarithms and antilog). Your last explanation was very clear and i even understood Law of tangent a little better.
my question was:
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".
give check
You helped me find angle A and B by using logarithms.
A=80 deg 26' 36"
B=67 deg 04' 55"
C=32 deg 28' 30"
a= 21.46
b= 46.28
and I used law of sine to find c.
this is my answer and i would appreciate if you correct me if I'm wrong.
sinA/a = sinC/c
sin 80.443 /21.46 = sin 32.475/c
21.46 (sin 32.475)/(sin 80.443)c
c= 11.68
okay now it says to give check.
okay i honestly don't know how or if the law of sin already covers the check part. I'd like your advice.
and thank you again . I honestly love it when i understand something i thought was so confusingly impossible to understand before.
4 answers
I have looked at the question, and confirm that there was an error in the previous calculations using the law of tangents. Following are results using the sine rule:
Given:
a = 21.46, b = 46.28, C = 32-28-30
The cosine rule gives:
c = sqrt(a^2+b^2-2*a*b*cos(C))=30.440833
The sine rule gives
A=22-14-32, and
B=125-16-58
If you draw the triangle, with C=32°, a=21.46 and b=46.28, you will see a skew triangle which shows obviously that B>90°.
So sin(A) has to be interpreted as 180 - arcsin() of the acute angle, which gives 125-16-58 (instead of 54-43-02 straight from the calculator).
I have not completed looking into the source of the error of my previous calculations. It may have to do with the interpretation of the atan() values. I will get back to you when it is done.
I apologize for the inconvenience.
Given:
a = 21.46, b = 46.28, C = 32-28-30
The cosine rule gives:
c = sqrt(a^2+b^2-2*a*b*cos(C))=30.440833
The sine rule gives
A=22-14-32, and
B=125-16-58
If you draw the triangle, with C=32°, a=21.46 and b=46.28, you will see a skew triangle which shows obviously that B>90°.
So sin(A) has to be interpreted as 180 - arcsin() of the acute angle, which gives 125-16-58 (instead of 54-43-02 straight from the calculator).
I have not completed looking into the source of the error of my previous calculations. It may have to do with the interpretation of the atan() values. I will get back to you when it is done.
I apologize for the inconvenience.
I just took another look. The formula for the law of tangents should have been tan((A+B)/s)/tan((A-B)/2)=(a+b)/(a-b).
Previously I have not divided the angles by 2, hence the error.
I will try to make a corrected version and repost in the original post.
Previously I have not divided the angles by 2, hence the error.
I will try to make a corrected version and repost in the original post.
The correction to the problem using the tangent rule has been posted at the original post:
http://www.jiskha.com/display.cgi?id=1298609593
Sorry for the inconvenience so caused.
http://www.jiskha.com/display.cgi?id=1298609593
Sorry for the inconvenience so caused.
3 answers