SOLUTION: Hi, I am Arun. I have problem in solving triangles. In a triangle ABC, AC is 10, BC is 0.025 and AB is 9.978. And the angle between BCA is said to be 30 degree. I need to find th

Algebra ->  Triangles -> SOLUTION: Hi, I am Arun. I have problem in solving triangles. In a triangle ABC, AC is 10, BC is 0.025 and AB is 9.978. And the angle between BCA is said to be 30 degree. I need to find th      Log On


   

Question 933855: Hi,
I am Arun. I have problem in solving triangles. In a triangle ABC, AC is 10, BC is 0.025 and AB is 9.978. And the angle between BCA is said to be 30 degree. I need to find the other two angles. I have used the cosine law of triangles and many other methods. For each method, i am getting a different solution for these two angles CAB and CBA. Can you help me in solving this?.
Thank you. Waiting for your response.
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
In a triangle ABC, AC is 10, BC is 0.025 and AB is 9.978. And the angle between BCA is said to be 30 degree. I need to find the other two angles. I have used the cosine law of triangles and many other methods. For each method, i am getting a different solution for these two angles CAB and CBA. Can you help me in solving this?.
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Using the given sides, BCA is not 30 degs, it's ~ 88.8494 degs
Use the Law of Sines to find the other angles.
Angle B = 5.738 degs
Angle A = 85.413 degs

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
check your side a; with given a=0.025 is impossible to get real solution for angle A using law of cos:
a%5E2=c%5E2%2Bb%5E2-2bc%2Acos%28A%29
%280.025%29%5E2=10%5E2%2B%289.978%29%5E2-%2810%29%289.978%29%2Acos%28A%29
0.000625=100%2B99.560484-99.78%2Acos%28A%29
0.000625=199.560484-99.78%2Acos%28A%29
99.78%2Acos%28A%29=199.560484-0.000625
99.78%2Acos%28A%29=199.559859
cos%28A%29=199.559859%2F99.78
cos%28A%29=1.9999986

A=1.3169571%2Ai(result in radians)
A=75.46%2Ai ° or A=0.0716197° which is very small angle

I was trying to find angle B using side b=9.978, side c=10, and angle C=30 using law of sin and got these results:

a+=+17.3+ you can get angle C+=+30°
b+=+9.98
c+=+10
Angles:
A+=+120°
B+=+29.9°
C+=+30°
so, IF given angle is 30, then a+=+17.3+
or, if you trying to find angle A using side a=0.025, side c=10, and angle C=30%7D%7D%2C+then+side+%7B%7B%7Bb must be 10.0216
results are as following:
Sides:
a+=+0.025+
b+=+10.0216+
c+=+10+
Angles:
A+=+0.0716197°......I wouldn't prefer this solution because the angle is very, very small
B+=+149.928°
C+=+30°
when you got right sides length, would be possible to find real solutions for angles