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

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

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Answer 1 · 2018-03-25T10:11:26

$c = 0, 9109, ˆ A = {23.65}^{\circ}, ˆ B = {53.35}^{\circ}$ $color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@$

$Area of Triangle A_{t} = (\frac{1}{2}) a b sin C = 0.137 sq units$ $color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = 0.137 "sq units"$

Explanation:

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

$ˆ C = 103^{\circ}, a = \frac{3}{8}, b = \frac{3}{4}, To solve the triangle$ $hat C = 103^@, a = 3/8, b = 3/4, " To solve the triangle"$

Applying Law of cosines,

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

$c = \sqrt{{(\frac{3}{8})}^{2} + {(\frac{3}{4})}^{2} - (2 \cdot (\frac{3}{8}) \cdot (\frac{3}{4}) \cdot cos 103)}$ $c = sqrt((3/8)^2 + (3/4)^2 - (2 * (3/8) * (3/4) * cos 103))$

$c = 0.9109$ $c = 0.9109$

Applying Law of sines,

$\frac{sin A}{\frac{3}{8}} = \frac{sin B}{\frac{3}{4}} = \frac{sin 103}{0.9109}$ $sin A / (3/8) = sin B / (3/4) = sin 103 / 0.9109$

$sin A = \frac{(\frac{3}{8}) \cdot sin 103}{0.9109} = 0.4011$ $sin A = ((3/8) * sin 103) / 0.9109 = 0.4011$

$ˆ A = {sin}^{- 1} 0.4011 = {23.65}^{\circ}$ $hat A = sin ^-1 0.4011 = 23.65^@$

Similarly, $sin B = \frac{(\frac{3}{4}) \cdot sin 103}{0.9109} = 0.8023$ $sin B = ((3/4) * sin 103) / 0.9109 = 0.8023$

$ˆ A = {sin}^{- 1} 0.8023 = {53.35}^{\circ}$ $hat A = sin ^-1 0.8023 = 53.35^@$

$Area of Triangle A_{t} = (\frac{1}{2}) a b sin C = (\frac{1}{2}) \cdot (\frac{3}{8}) \cdot (\frac{3}{4}) \cdot sin 103 = 0.137 sq units$ $color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 "sq units"$

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

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