How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 131^{\circ}, a = 15, b = 32$ $A=131^circ, a=15, b=32$ ?

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 131^{\circ}, a = 15, b = 32$ $A=131^circ, a=15, b=32$ ?

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Answer 1 · 2018-07-15T10:23:37

No solution with the given measurements.

Explanation:

$ˆ A = 131^{\circ}, a = 15, b = 32$ $hat A = 131^@, a = 15, b = 32$

$since side b is greater than side a, ˆ B is to be greater than ˆ A$ $"Since side " b " is greater than side " a, hat B " is to be greater than " hat A$

$But ' ˆ A = 131^{\circ}, and hence ˆ B to be greater than 131^{\circ}$ $"But '" hat A = 131^@, " and hence " hat B " to be greater than " 131^@$

$since sum of the three angles of a triangle is to be 180^{\circ}, we cannot form a triangle with the given measurements$ $"Since sum of the three angles of a triangle is to be " 180^@, " we cannot form a triangle with the given measurements"$

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 131^{\circ}, a = 15, b = 32$ $A=131^circ, a=15, b=32$ ?

1 Answer

Explanation:

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How do you determine whether △ABCtriangle ABC has no, one, or two solutions given A=131∘,a=15,b=32A=131^circ, a=15, b=32?

1 Answer

Explanation:

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 131^{\circ}, a = 15, b = 32$ $A=131^circ, a=15, b=32$ ?