How do you solve the triangle given $C=103^circ, a=3/8, b=3/4$ ?

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Answer 1 · 2018-03-25T10:11:26

$color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@$

$color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = 0.137 "sq units"$

![http://www.dummies.com/education/math/trigonometry/laws-of-sines-and-cosines/]( useruploads.socratic.org )

$hat C = 103^@, a = 3/8, b = 3/4, " To solve the triangle"$

Applying Law of cosines,

$c = sqrt(a^2 + b^2 - 2 a b cos C)$

$c = sqrt((3/8)^2 + (3/4)^2 - (2 * (3/8) * (3/4) * cos 103))$

$c = 0.9109$

Applying Law of sines,

$sin A / (3/8) = sin B / (3/4) = sin 103 / 0.9109$

$sin A = ((3/8) * sin 103) / 0.9109 = 0.4011$

$hat A = sin ^-1 0.4011 = 23.65^@$

Similarly, $sin B = ((3/4) * sin 103) / 0.9109 = 0.8023$

$hat A = sin ^-1 0.8023 = 53.35^@$

$color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 "sq units"$

How do you solve the triangle given C=103^circ, a=3/8, b=3/4C=103^circ, a=3/8, b=3/4?