How do you use law of sines to solve the triangle given A = 57° , a = 11, b = 10?

1 Answer
Nov 23, 2015

A=57°
B=arcsin((10sin(57°))/11)~~49.678696410324°
C=123°-arcsin((10sin(57°))/11)~~73.321303589676°
a=11
b=10
c=(11sin(123°-arcsin((10sin(57°))/11)))/sin(57°)~~12.564196743012485

Explanation:

First, you should know what the Law of Sines is. It is simply this:

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

Now let's look at our given. We have:

A=57°, a=11, and b=10

We need to look for B, C, and c.

Step 1 - Solving for B

Law of Sines

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

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

[2]" "11/sin(57°)=10/sinB

Multiply both sides by (sinB)[sin(57°)].

[3]" "11/cancelsin(57°)(sinB)cancel[sin(57°)]=10/cancelsinBcancel(sinB)[sin(57°)]

[4]" "11sinB=10sin(57°)

Divide both sides by 11.

[5]" "sinB=(10sin(57°))/11

Apply arcsin on both sides.

[6]" "arcsin(sinB)=arcsin((10sin(57°))/11)

[7]" "color(blue)(B=arcsin((10sin(57°))/11)~~49.678696410324°)

Step 2 - Solving for C

The sum of the interior angles of all triangles 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°-57°-arcsin((10sin(57°))/11)

[4]" "color(blue)(C=123°-arcsin((10sin(57°))/11)~~73.321303589676°)

Step 3 - Solving for c

Law of Sines

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

Multiply both sides by sinC

[2]" "c/cancelsinCcancel(sinC)=a/sinA(sinC)

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

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

[4]" "color(blue)(c=(11sin(123°-arcsin((10sin(57°))/11)))/sin(57°)~~12.564196743012485)