SOLUTION: Here is the image:http://i.imgur.com/L2ZLKEM.png
a=38, c=44, ∠A = 35°
What I have so far:
sin(35)/38 = sinC/44
44sin(35) = 38sinC
44sin(35)/38 = sinC
sinC
Algebra.Com
Question 932322: Here is the image:http://i.imgur.com/L2ZLKEM.png
a=38, c=44, ∠A = 35°
What I have so far:
sin(35)/38 = sinC/44
44sin(35) = 38sinC
44sin(35)/38 = sinC
sinC = 0.6641
sin^-1(0.6641)
∠C = 41.6º
Is this ∠C1?
After this I don't know how to derivive the rest.
---------------------------------------------
For the second problem:
a = 76, b = 105, ∠A = 29°
sin(29)/76 = sinB/105
105sin(29) = 76sinB
sinB = 105sin(29)/76
sin^-1(0.6698)
∠B = 42.1º
Again, after this point I am lost
I could really use some instruction on how to solve.
Thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a=38, c=44, ∠A = 35°
What I have so far:
sin(35)/38 = sinC/44
44sin(35) = 38sinC
44sin(35)/38 = sinC
sinC = 0.6641
sin^-1(0.6641)
∠C = 41.6º
Is this ∠C1?
Comment:: yes
---
B = 180-(35+41.6) = 103.39 degrees
----
c/sin(C) = b/sin(B)
44/0.6641 = b/0.97
b = 0.97*44/0.6641
b = 64.45
----------------
Cheers,
Stan H.
-------------------
----------------------------------------------
After this I don't know how to derive the rest.
Comment:: After you have A and B, subtract their sum from 180 to get C
Use the Law of Sines to find side "c".
---------------------------------------------
For the second problem:
a = 76, b = 105, ∠A = 29°
sin(29)/76 = sinB/105
105sin(29) = 76sinB
sinB = 105sin(29)/76
sin^-1(0.6698)
∠B = 42.1º
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