How do you solve the triangle when Angle A=37 degrees, a=49, b=54?

1 Answer
Nov 24, 2015

#A=37°#
#a=49#
#B=arcsin((54sin(37°))/49)~~41.546274420887°#
#b=54#
#C=143°-arcsin((54sin(37°))/49)~~101.453725579113°#
#c=(49sin(143°-arcsin((54sin(37°))/49)))/sin(37°)~~79.7989130980759269#

Explanation:

Make use of the Law of Sines:

#a/sinA=b/sinB=c/sinC#

Solving for #B#

Law of Sines

#[1]" "a/sinA=b/sinB#

Isolate #sinB#.

#[2]" "cancela/cancelsinA*(cancel((sinA))(sinB))/cancela=b/cancelsinB*((sinA)cancel((sinB)))/a#

#[3]" "sinB=(bsinA)/a#

Plug in the values of #a#, #A#, and #b#.

#[4]" "sinB=(54sin(37°))/49#

Get the inverse function. Solve using a calculator

#[5]" "hArrcolor(blue)(B=arcsin((54sin(37°))/49)~~41.546274420887°)#

Solving for #C#

The sum of all interior angles of a triangle is #180°#.

#[1]" "A+B+C=180°#

Isolate #C#

#[2]" "C=180°-A-B#

Plug in the values of #A# and #B#.

#[3]" "C=180°-37°-arcsin((54sin(37°))/49)#

Solve using a calculator.

#[4]" "color(blue)(C=143°-arcsin((54sin(37°))/49)~~101.453725579113°)#

Solving for #c#

Law of Sines

#[1]" "c/sinC=a/sinA#

Isolate #c#.

#[2]" "c/cancelsinC*cancelsinC=a/sinA*sinC#

#[3]" "c=(asinC)/sinA#

Plug in the values of #A#, #a#, and #C#.

#[4]" "c=(49sin(143°-arcsin((54sin(37°))/49)))/sin(37°)#

Solve using a calculator.

#[5]" "color(blue)(c=(49sin(143°-arcsin((54sin(37°))/49)))/sin(37°)~~79.7989130980759269)#