March 9th, 2017, 02:38 PM  #1 
Newbie Joined: Mar 2017 From: Michigan Posts: 3 Thanks: 0  Finding The Length Of AB
I'm currently having trouble with this question, my teacher has given me. Find in triangle ABC, sinA=0.5, sinB=0.7, a=9, Find the length of AB. The triangle looks like the one present. Thanks in Advance! Last edited by TotalMathMan; March 9th, 2017 at 03:19 PM. 
March 9th, 2017, 03:08 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709 
SinC cannot equal 9 you're going to need the length of at least one side. that triangle can be grown or shrunk as much as you like and still keep those angles. Check the problem again. 
March 9th, 2017, 03:11 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364 
I think that's a trick question. Do you know about similar triangles? There's a triangle with those same angles for any length you can name. 10 miles, a million miles, an inch. Did you leave out any info? Or does your teacher like to play tricks? 
March 9th, 2017, 03:16 PM  #4  
Newbie Joined: Mar 2017 From: Michigan Posts: 3 Thanks: 0  Quote:
I went ahead and adjust it I believe I might have read it wrong, thank you for pointing that out.  
March 9th, 2017, 03:18 PM  #5  
Newbie Joined: Mar 2017 From: Michigan Posts: 3 Thanks: 0  Quote:
I doubt it, worse case scenario I can go ahead and ask for help tomorrow from her.  
March 9th, 2017, 04:50 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,908 Thanks: 1382 
Which of the triangle's angles is obtuse?

March 9th, 2017, 06:06 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709  Just use the law of sines; it's a one line calculation.
Last edited by skipjack; March 10th, 2017 at 03:08 AM. 
March 10th, 2017, 03:09 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 17,908 Thanks: 1382 
It's not quite that simple, as the angle opposite AB isn't given.

March 10th, 2017, 06:19 AM  #9 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709  ok, not quite one line. Law of sines $\dfrac{\sin(A)}{a} = \dfrac{\sin(B)}{b}$ $\dfrac{0.5}{9} = \dfrac{0.7}{b}$ $b = 12.6$ $\sin(C) = \sin( \pi  \arcsin(0.5)  \arcsin(0.7) ) \approx 0.963$ $\dfrac{\sin(C)}{c} = \dfrac{\sin(A)}{a}$ $c \approx 17.34$ 
March 10th, 2017, 08:11 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,908 Thanks: 1382 
Why did you calculate the value of b, but then not use it to find c?


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