Question

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

Answer 1

`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)`