Pleasee Help!! If a triangle with a = 10.9 m, c = 30.2 m, and C = 50° exists. If it does, determine A to the nearest degree.?
2016-04-20 4:51 pm
回答 (3)
2016-04-20 5:00 pm
✔ 最佳答案
Use the Sine Rule Which is
SinA/a = SinB/b = SinC/c
Permutate any two of the above out of the three.
Hence
SinA/a = SinC / c =>
SinA = a*SinC/c
SinA = 10.9 (Sin50) / 30.2
SinA = 10.9(0.766044443) / 30.2
SinA = 0.276486239...
A = Sin^-1(0.2764...)
A = 16.05 degrees.
A = 16 degrees ( nearest degree).
2016-04-20 5:10 pm
a/sin A = b/sin B = c/sin C
10.9/sin A = c/sin(50)
=> sin A = 10.9 sin(50) /30.2
sin A = 0.276
A = sin^-1(0.276)
A = 16.05°
B = 180 - (50+16.05) = 113.95°
b/sin B = c/sin C
b/sin(113.95) = 30.2/sin(50)
b = sin(113.95) *30.2 /sin(50)
= 36.03 m
10.9/sin A = c/sin(50)
=> sin A = 10.9 sin(50) /30.2
sin A = 0.276
A = sin^-1(0.276)
A = 16.05°
B = 180 - (50+16.05) = 113.95°
b/sin B = c/sin C
b/sin(113.95) = 30.2/sin(50)
b = sin(113.95) *30.2 /sin(50)
= 36.03 m
收錄日期: 2021-04-28 23:54:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160420085133AAdr1c8