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 January 31st, 2008, 10:06 PM #1 Newbie   Joined: Jan 2008 Posts: 2 Thanks: 0 Sine Rule Consider the triangle ABC where A=30 degrees b=10 and AB is the horizontal base of the triangle. I have a problem trying to find the smallest value of "a" for which the triangle ABC can exist and range of values of a for two triangles to exist. I calculated B=sin^-1 [10 sin (30)]/a I can continue plugging in numbers to find the smallest value, but is there a better way? And I really don't know how to find value for two triangles. Last edited by skipjack; February 23rd, 2018 at 08:32 PM.
February 1st, 2008, 05:07 AM   #2
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Quote:
 Originally Posted by Jenty850 I calculated B=sin^-1 [10 sin (30)]/a.
You should have obtained B = sin^-1[10sin(30°)/a] (or 180° - sin^-1[10sin(30°)/a]), which requires 10sin(30°)/a ≤ 1, i.e., 5 ≤ a. There are two possible triangles if 5 < a < 10 (use a sketch to see why).

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