SOLUTION: Solve the triangle using Law of sines or cosines. A=42(angle A) b=120 (side b) c=160 (side c)

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Question 141297This question is from textbook Algebra 2
: Solve the triangle using Law of sines or cosines.
A=42(angle A)
b=120 (side b)
c=160 (side c)
This question is from textbook Algebra 2

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the triangle using Law of sines or cosines.
A=42(angle A)
b=120 (side b)
c=160 (side c)

Keep all decimals until the end, then round.

a%5E2=b%5E2%2Bc%5E2-2%2Ab%2Ac%2AcosA
a%5E2=120%5E2%2B160%5E2-2%2A120%2A160%2Acos42°

Make sure your calculator is in DEGREE mode
and not RADIAN mode.

Punch that in your calculator and you get:

a%5E2=11463.2387

Then taking square roots of both sides

a+=+107.0665153

Now we use the law of sines:

a%2FsinA=b%2FsinB=c%2FsinC

We need only the first two parts:

a%2FsinA=b%2FsinB

cross-multiply:

a%2AsinB+=+b%2AsinA

Divide both sides by a

sinB+=+%28b%2AsinA%29%2Fa

Substitute:

sinB+=+%28120%2Asin42%29%2F107.0665153

Calculate

sinB+=+0.7499606439

No use the inverse sine key 
on your calculator

B+=+48.58696886°

Now we find angle C from

A+%2B+B+%2B+C+=+180°

C+=+180+-+A+-+B

C+=+180+-+42+-+48.58696886

C+=+180+-+42+-+48.58696886

C+=+89.41303114°
  
So the three missing parts are

a+=+107.0665153
B+=+48.58696886°
C+=+89.41303114°

But using the rounding rules, the angles
should be rounded to the nearest degree just 
like the given angle 42°, and the side should
be rounded to the nearest two significant 
digits.  So using the rounding rules:

a+=+110  (side a)
B+=+49°  (Angle B) 
C+=+89°  (Angle C)

Edwin