Answer on Question #51011 – Math – Trigonometry
Solve triangle ABC which has angle C=1250:431; a=4:2 cm and c=8:2cm. Find b
Solution
Fig.1
The law of cosines (see Fig.1) tells us that
c2  a2  b2  2ab cos C
which is equivalent to
b2  2ab cos C  a2  c2  0
After plugging all known values, obtain the following equation:
2
2
4
1250  4   8 
b  2  b cos
     0
2
431  2   2 
2
that is,
b2  4b cos
2

 1250  
D   4 cos 
   4 12
 431  

2
  1250  
 1250 
4 cos 
  16 cos 
   4 12
 431 
  431  
b1,2 
2
Because b  0 as the length of triangle, then take
1250
 12  0
431
2
  1250  
 1250 
4 cos 
2
  16 cos 
   4 12
  1250  
 431 
 1250 
  431  
b
 2 cos 
  2 cos 
   3  2.03cm
2
 431 
  431  
2
1250 
  1250  
Answer: b  2cos 
  2 cos 
   3  2.03cm
 431 
  431  
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