Question

In the following triangle, theta=60 degrees. Find the values of B and B', which solve this ambiguous case.

Please explain. Thank you!

Answer 1

Use the inverse of sine to find the angle that's not theta or b' the other one so you can find b' you know how to do that right?

Answer 2

And then there triangle with the angle b is an isosceles triangle so b and the angle next to it are the same

Answer 3

Ok I am going to try right now.

Answer 4

HI!!

Answer 5

this is funny because you really don't have to do any more work once you get B

Answer 6

only one answer has the correct B, so B' is not really even part of the question is it?

Answer 7

that is the beauty of multiple choice questions
really a lazy way to do math for sure

Answer 8

So the other angle not b' and not theta is 66.9 then add 60 to get 126.9 degrees then subtract from 189 to get 53.1 degrees

Answer 9

9.2/sin 60 = 10/sin B
sin B=sin60*10/9.2=0.941331961
B=sin^-1 0.941331961=70.27 degrees
B'=180-70.27=109.7235415 or 109.73 degrees

Answer 10

Answer C. I guess?

Answer 11

yes, easier to just use this
http://www.wolframalpha.com/input/?i=arcsin%2810sin%2860%29%2F9.2%29

Answer 12

and as always you get C

Answer 13

but really you should try to do it in one step, rather than computing all along the way
\[B=\sin^{-1}(\frac{10\sin(60)}{9.2})\]

Answer 14

ok that looks a lot easier :)

Answer 15

Use the law of sines:

\(\dfrac{9.2}{\sin \theta} = \dfrac{10}{\sin B} \)

\(\dfrac{9.2}{\sin 60^o} = \dfrac{10}{\sin B} \)

\(\sin B = \dfrac{10\sin 60^o}{9.2} \)

\(B = \sin^{-1} \dfrac{10\sin 60^o}{9.2} \)