Question

Can someone help me with this "ambiguous" triangle? Will give medal & fan.

"In the following triangle, find the values of the angles B and B', which are the best approximations to the solutions to this ambiguous case?"
A. B = 70.5 or B' = 109.5
B. B = 71.6 or B' = 108.4
C. B = 70 or B' = 122.5
D. B = 67 or B' = 113

Answers are in degrees.

Answer 1

You are using Law of Sines, right? So you have:

sin45*/13.5 = sinB'/18
sinB'=18(sin45*/13.5)

So that gives you a value for sinB'

Now use the inverse sine function to get B'. That will be the first quandrant (ie., the acute angle) solution.

Then the other solution is just 180* - the first solution.

Answer 2

Yes I am using Law of Sines. So wait, I take the inverse sine for B', so it would be
sin^-1 (18(sin45/13.5))? Or about 3?

Answer 3

That doesn't seem right to me ): I think I'm doing it wrong

Answer 4

@DebbieG ?

Answer 5

@Hero or @phi ? Can either of you help please?

Answer 6

make sure your calculator is in degree mode.

Answer 7

How do I do that? I'm using my laptop's calculator in scientific mode.

Answer 8

first, what do you get for this part
18*sin(45)/13.5

Answer 9

0.9428...

Answer 10

That's correct. But you should not be getting 3 for your answer.... in either degree or radian mode.

Answer 11

ok that is good

Answer 12

now do sin^-1 on 0.9428

Answer 13

okay now I got 70.5, not sure what I did wrong before lol

Answer 14

yes.

the other angle in the choice should give you the same sin 0.9428

Answer 15

So wait I'm confused is 70.5 B' or B?

Answer 16

sin 109.5= 0.9426 which is close (they rounded it)

If you look at the picture, one of the angles is less than 90
and one is bigger than 90

Answer 17

Oh I get it, so the answer would be A? B = 70.5 and B' = 109.5?

Answer 18

yes

Answer 19

Thank you so much!

Answer 20

yw. Thank DebbieG also.

Answer 21

Thank you both!

Answer 22

Do you understand why? For any positive sine value (e.g., any real number between 0 and 1), there is an acute angle (in the first quadrant) and also an obtuse angle (in the 2nd quadrant) that has that sine value. So each of those is a solution to this triangle - that's what makes it the "ambiguous case", you can't pin it down to just one triangle. :)

Answer 23

Sure thing! Good teamwork. :)

Answer 24

Ohh okay, yes I completely understand now (: I'm writing that in my notes, thanks again