Greenrune113 Greenrune113
  • 17-05-2018
  • Mathematics
contestada

QF Q2.) Solve the triangle

QF Q2 Solve the triangle class=

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Asdfjkadfjkakjf Asdfjkadfjkakjf
  • 19-05-2018
C=180-A-B=180-67-53=60deg.

by sine rule

b/sinB=a/sinA
b=a/sinA*sinB
=13/sin(67)*sin(53)
= 11.28

c/sinC=a/sinA
c=a/sinA*sinC
=13/sin(67)*sin(60)
= 12.23
Answer Link
nustopocru nustopocru
  • 19-05-2018

Sum of interior angles in a triangle is 180 deg.

so angle C = 180 - A - B

=180 - 67 - 53

=60deg.


Use the sine rule in a triangle - sinA / a = sinB / b


b = a * sinB / sinA
=13 * sin(53) / sin(67)
=11.3

Similarly sinC / c = sinA / a
c = a * sinC / sinA
=13 * sin(60) / sin(67)
=12.2


Answer Link

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