How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 52^{\circ}, a = 9, b = 20$ $A=52^circ, a=9, b=20$ ?

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Answer 1 · 2018-06-21T07:16:14

$As sin B can have a value only between - 1 & + 1$ $color(crimson)("As "sin B " can have a value only between " -1 " & "+1$ ,

$Given sum doesn't have a solution$ $color(blue)("Given sum doesn't have a solution"$

![https://sites.google.com/site/trigingwithstarr/law-of-sines-and-cosines]( useruploads.socratic.org )

$ˆ A = 52^{\circ}, a = 9, b = 20$ $hat A = 52^@, a = 9, b = 20$

Applying Law of Sines,

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a / sin A = b / sin B = c / sin C$

$sin B = \frac{b sin A}{a} = \frac{20 \cdot sin 52}{9} = 1.7511$ $sin B = (b sin A ) / a = (20 * sin 52 ) / 9 = 1.7511$

$As sin B can have a value only between - 1 & + 1$ $"As "sin B " can have a value only between " -1 " & "+1$ ,

$Given sum doesn't have a solution$ $"Given sum doesn't have a solution"$

How do you determine whether △ABCtriangle ABC has no, one, or two solutions given A=52∘,a=9,b=20A=52^circ, a=9, b=20?