How do you solve a triangle given A=58 degrees, a=11.4, b=12.8?

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How do you solve a triangle given A=58 degrees, a=11.4, b=12.8?

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Answer 1 · 2018-05-24T02:51:09

Case 1 : $ˆ A \equiv 58, ˆ B = 72.2, ˆ C = 49.8, a = 11.4, b = 12.8, c = 10.27$ $color(blue)(hat A -= 58, hat B = 72.2, hat C = 49.8, a = 11.4, b = 12.8, c = 10.27$

Case 2 $ˆ A = 58, ˆ B = 106.8, ˆ C = 14.2, a = 11.4, b = 12.8, c = 3.3$ $color(green)(hat A = 58, hat B = 106.8, hat C = 14.2, a = 11.4, b = 12.8, c = 3.3$

Explanation:

$a = 11.4, b = 12.8, ˆ A = 58^{\circ}$ $a = 11.4, b = 12.8, hat A = 58^@$

Applying Law of Sines,

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

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

$sin B = \frac{12.8 \cdot sin 58}{11.4} = 0.9522$ $sin B = (12.8 * sin 58)/ 11.4 = 0.9522$

$ˆ B = {sin}^{- 1} 0.9522 = {72.2}^{\circ}$ $hat B = sin ^ -1 0.9522 = 72.2^@$

Also sin B = sin (180 - B)#

Hence $ˆ B = 180 - 72.2 = {107.8}^{\circ}$ $hat B = 180-72.2 = 107.8^@$

$ˆ C = 180 - 58 - 72.2 = {49.8}^{\circ}$ $hat C = 180 - 58 - 72.2 = 49.8^@$

Also $ˆ C = 180 - 58 - 107.8 = {14.2}^{\circ}$ $hat C = 180 - 58 - 107.8 = 14.2^@$

$c = \frac{a \cdot sin C}{sin A} = \frac{11.4 \cdot sin 49.8}{sin 58}$ $c = (a * sin C) / sin A = (11.4 * sin 49.8) / sin 58$

$c = 10.27$ $c = 10.27$

Also $c = \frac{11.4 \cdot sin 14.2}{sin 58} = 3.3$ $c = (11.4*sin 14.2)/sin 58 = 3.3$

Measurements of triangle are

Case 1 : $ˆ A \equiv 58, ˆ B = 72.2, ˆ C = 49.8, a = 11.4, b = 12.8, c = 10.27$ $color(blue)(hat A -= 58, hat B = 72.2, hat C = 49.8, a = 11.4, b = 12.8, c = 10.27$

Case 2 $ˆ A = 58, ˆ B = 106.8, ˆ C = 14.2, a = 11.4, b = 12.8, c = 3.3$ $color(green)(hat A = 58, hat B = 106.8, hat C = 14.2, a = 11.4, b = 12.8, c = 3.3$

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How do you solve a triangle given A=58 degrees, a=11.4, b=12.8?

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