How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 34^{\circ}, a = 8, b = 13$ $A=34^circ, a=8, b=13$ ?

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Answer 1 · 2017-11-21T15:15:51

$2$ $2$ triangles.

If

$a > b$ $a > b$

then one triangle is formed, but if

$a < b$ $a < b$

then proceed to the formula

$a < = > b sin (A)$ $a " < = > " bsin(A)$

In which

So in the problem

$8 <=> 13 sin (34)$ $8 " <=>" 13sin(34)$

$8 > 7.27$ $8>7.27$

Therefore, two triangles will be formed.

Or you could use the other way (sin law)

$\frac{a}{sin (A)} = \frac{b}{sin (B)}$ $a/sin(A)=b/sin(B)$

$\frac{8}{sin (34)} = \frac{13}{sin (x)}$ $8/sin(34)=13/sin(x)$

$8 sin (x) = 13 sin (34)$ $8sin(x)=13sin(34)$

$sin (x) = \frac{13 sin (34)}{8}$ $sin(x)=(13sin(34))/8$

$x = {sin}^{- 1} (\frac{13 sin (34)}{8})$ $x=sin^-1((13sin(34))/8)$

$△ 1 = 65.32$ $triangle1=65.32$
$△ 2 = 180 - 65.32 = 114.68$ $triangle2=180-65.32=114.68$

Proving

$65.32 + 34 < 180$ $65.32+34<180$ $\sqrt{}$ $sqrt$
$114.68 + 34 < 180$ $114.68+34<180$ $\sqrt{}$ $sqrt$

How do you determine whether △ABCtriangle ABC has no, one, or two solutions given A=34∘,a=8,b=13A=34^circ, a=8, b=13?