How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 100^{\circ}, a = 7, b = 3$ $A=100^circ, a=7, b=3$ ?

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Answer 1 · 2018-02-19T06:36:27

One solution : $a = 7, b = 3, c = 5.82$ $color(green)(a = 7, b = 3, c = 5.82$

$ˆ A = 100^{\circ}, ˆ B = 25^{\circ}, ˆ C = 55^{\circ}$ $color(blue)(hatA = 100^@, hatB = 25^@, hatC = 55^@$

Given $ˆ A = 100^{\circ}, a = 7, b = 3$ $hatA = 100^@, a = 7, b = 3$

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As per sine law, $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a / sin A = b / sin B = c / sin C$

$sin B = \frac{b \cdot sin A}{a} = \frac{3 \cdot sin 100}{7} = 0.4221$ $sin B = (b * sin A ) / a = (3 * sin 100) / 7 = 0.4221$

$ˆ B = {sin}^{- 1} 0.4221 \approx 25^{\circ}$ $hatB = sin ^(-1) 0.4221 ~~ 25^@$

$ˆ C = 180 - 100 - 25 = 55^{\circ}$ $hatC = 180 - 100 - 25 = 55^@$

$c = \frac{a \cdot sin C}{sin A} = \frac{7 \cdot sin 55}{sin 100} \approx 5.82$ $c = (a * sin C) / sin A = (7 * sin 55) / sin 100 ~~ 5.82$

How do you determine whether △ABCtriangle ABC has no, one, or two solutions given A=100∘,a=7,b=3A=100^circ, a=7, b=3?