Question

Solve the triangle.

A = 21°, C = 105°, c = 5

Answer 1

so.... what are we solving for?

Answer 2

the other components.
Can you check my solution?
B = 54°, a ≈ 1.9, b ≈ 4.2

Answer 3

You can easily solve for B, first
180=a+b+c so 180-105-21=54

Then you can use law of sines for this one
\[\frac{ sinA }{ a }=\frac{ sinB }{ b }\frac{ sinC }{ c }\]
Then set it up with the information you know
\[\frac{ \sin21^{o} }{ a }=\frac{ sinB }{ b } = \frac{ \sin105^{o} }{ 5 }\]

You can basically ignore the middle part about sinB and b for now, and take the a and c parts and cross multiply, solve for a
\[5\sin21^{o}=asin105^{o}\]
Divide both sides by sin105, and a= 1.86 (MAKE SURE YOURE IN DEGREES NOT RADIANS)
Then do the same for b with the cross multiplying
\[5\sin54^{o}=bsin105\]
Divide by sin105, b=4.19

Then its always good to make sure your biggest angle and biggest side line up so that you know you didn't do something stupid

Answer 4

So, was I correct?

Answer 5

Haha yes, you sent your answers before i got a chance to finish. Good work! :)

Answer 6

Can you help me with something else?

Answer 7

you can always post anew, thus we can revise each other