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

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Answer 1 · 2017-11-21T15:15:51

#2# triangles.

If

#a > b#

then one triangle is formed, but if

#a < b#

then proceed to the formula

#a " < = > " bsin(A)#

In which

So in the problem

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

#8>7.27#

Therefore, two triangles will be formed.

Or you could use the other way (sin law)

#a/sin(A)=b/sin(B)#

#8/sin(34)=13/sin(x)#

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

#sin(x)=(13sin(34))/8#

#x=sin^-1((13sin(34))/8)#

#triangle1=65.32#
#triangle2=180-65.32=114.68#

Proving

#65.32+34<180# #sqrt#
#114.68+34<180# #sqrt#