A triangle has sides A, B, and C. The angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 72, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 72, what is the area of the triangle?

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Answer 1 · 2018-06-03T09:48:24

Area of the triangle $A_{t} = 748.32$ $A_t = color(crimson)(748.32)$

Explanation:

$ˆ A = \frac{π}{12}, ˆ C = (7 π) : 12, ˆ B = \frac{π}{3}, b = 72$ $hat A = pi/12,hat C = (7pi):12, hat B = pi/3, b = 72$

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

$A r e a = A_{t} = (\frac{1}{2}) a b sin C$ $Area = A_t = (1/2) a b sin C$

$a = \frac{b sin A}{sin B} = \frac{72 \cdot sin (\frac{π}{12})}{sin (\frac{π}{3})} = 21.52$ $a = (b sin A) / sin B = (72 * sin (pi/12)) / sin (pi/3) = 21.52$

$A_{t} = (\frac{1}{2}) \cdot 21.52 \cdot 72 \cdot sin (\frac{7 π}{12}) = 748.32$ $A_t = (1/2) * 21.52*72*sin((7pi)/12) = color(crimson)(748.32)$

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 72, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 72, what is the area of the triangle?