How do you find the area of a triangle given A=18, B=35, c=3.4?

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Answer 1 · 2018-05-25T04:08:16

$A_{t} = 3.31$ $color(purple)(A_t = 3.31$ sq units

$ˆ A = 18^{\circ}, ˆ B = 35^{\circ}, c = 3.4$ $hat A = 18^@, hat B = 35^@, c = 3.4$

$ˆ C = (180 - 18 - 35) = 127^{\circ}$ $hat C = (180-18-35) = 127^@$

As per Law of Sines,

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

$b = \frac{c \cdot sin B}{sin C}$ $b = (c * sin B) / sin C$

$b = \frac{3.4 \cdot sin 35}{sin 127} = 2.44$ $b = (3.4 * sin 35) / sin 127 = 2.44$

Area of triangle - $A_{t} = (\frac{1}{2}) b c sin A$ $A_t = (1/2) b c sin A$

$A_{t} = (\frac{1}{2}) \cdot 2.44 \cdot 3.4 \cdot sin (127)$ $A_t = (1/2) * 2.44 * 3.4 * sin (127)$

$A_{t} = 3.31$ $color(purple)(A_t = 3.31$ sq units

How do you find the area of a triangle given A=18*, B=35*, c=3.4?