Find Sin B without the calculator, a=2, b=3, c=4?

Question

Sure this is not a right triangle
Sure my data is correct
Hint: answer is a single fraction with radical

Answer 1

Just wondering? Getting others to do your homework? I see at least a couple trig questions posted!

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While we may help...it is not helpful for you unless you are using our ideas to better understand the subject, and thus lose Yahoo Answers as your training wheels!

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Well, this isn't a right triangle, so you would be stuck with the Law of Cosines.


Double check, do you have your a, b and c correct?
....

hang on...

Okay, use the Law of Cosines, and find Cos B:

b^2 = a^2 + c^2 - 2ac Cos B

3^2 = 2^2 + 4^2 - 2(2)(4) Cos B

9 = 4 + 16 - 16 Cos B

9 = 20 - 16 Cos B

-11 = -16 Cos B

Cos B = 11/16

Now use the Pythagorean Identity:

Sin^2 B + Cos^2 B = 1

Sin^2 B + (11/16)^2 = 1

Sin^2 B = 1 - 121/256

Sin^2 B = 135/256

Sin B = sqrt(135) / 16

Hopefully that's right....

Answer 2

OK. b^2 = a^2 + c^2 - 2acCosB

So,
9=4+16-2*2*4*CosB

So, 16CosB=11 or CosB=11/16

And since Sin^2(B)+Cos^2(B)=1

So, Sin^2(B)+121/256=1

Which means that Sin^2(B)=135/256.

So, SinB = 3/16*SQRT(15)

Answer 3

B is the angle in front of side b. c is the hypotenuse.

Sin B = b/c = 3/4 or 0.75