Solve the following oblique triangles:
11. a = 134.2°, b = 84.54, B = 52° 9’ 11”
12. a = 627.7°, b = 412.2, A =66° 47
2007-01-03 14:39:32 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
plz do it.
2007-01-03 15:00:49 · update #1
11.sinA/a=sinB/b
sinA/134.2=sin52*9'11''/84.54
sinA=[(134.2/84.54)]*sin52*9'11''
once A is found C=180-(A+B)
and again using the sine formula side c can be found
12.627.7/sin 66*47'=412.2/sinB
sinB=(412.2/627.7)sin66*47'
once angle B is known C=180-(A+B)
once Cis found using the cosine formula
c^2=a^2+b^2-2abcosC, side c can be found
2007-01-03 15:32:07 · answer #1 · answered by raj 7 · 0⤊ 0⤋
11. a = 134.2°, b = 84.54, B = 52° 9’ 11”
- Solve for A using the law of sines.
sin A/a = sin B/b
- Then solve for the measurement of angle C by
m
- Then solve for c using the law of cosines.
c² = a² + b² - 2ab cos C
2007-01-03 22:45:30 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
Could you please clear up which are angles and which are sides, you're kinda confusing me! (how is 627 a degree measure in a triangle? i know i'm probably wrong about that somehow, but still :-)
2007-01-03 22:45:02 · answer #3 · answered by Panky1414 2 · 0⤊ 0⤋
I'll help you, just explain which letters are angles or sides measurements
2007-01-03 22:44:47 · answer #4 · answered by mzuleta2002 2 · 0⤊ 0⤋