Solve the triangle a = 3, b = 2 and alpha = 40 degrees

posader86 2022-07-30 Answered
Solve the triangle
a = 3, b = 2 and alpha = 40 degrees
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Answers (1)

dominicsheq8
Answered 2022-07-31 Author has 15 answers
a / sin ( A ) = b / sin ( B ) = c / sin ( C )
therefore a,b,c are the lengths of the sides
so 3 / sin ( 40 d e g ) = 2 / sin ( θ ) so arcsin ( 2 sin ( 40 ) / 3 ) = B or B=25.37deg
now every triangle is 90deg so 40deg+25.37deg=65.37deg so onlyangle left is 90-65.37=24.63degrees this is opposite of lengthc
so law of sines again using a and c now a / sin ( A ) = c / sin ( C )so 3 / sin ( 40 d e g ) = c / sin ( 24.63 d e g ) solve for c=1.945 units in length
the triangle's properties are...
a=3 b=4 c=1.945 A=40deg B=25.37deg C=24.63deg
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Find all triangles with a fixed base and opposite angle
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