A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{6}$ $(5pi)/6$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 2, what is the area of the triangle?

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Answer 1 · 2018-02-17T08:54:05

Area of triangle $A_{t} = (\frac{1}{2}) a b sin C = 1$ $color(blue)(A_t = (1/2) a b sin C = 1$

$G i v e n : b = 2, ˆ A = \frac{π}{12}, ˆ C = \frac{5 π}{6}$ $Given : b = 2, hatA = pi/12, hatC = (5pi)/6$

$ˆ B = π - \frac{π}{12} - \frac{5 π}{6} = \frac{π}{12}$ $hat B = pi - pi/12 - (5pi)/6 = pi/12$

It’s an isosceles triangle with $ˆ A = ˆ B$ $hatA = hatB$

$:. b = a = 2$

Area of triangle $A_t = (1/2) a b sin C = (1/2) * 2 * 2 * sin ((5pi)/6) = color(blue)(1)$

A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/65π6(5pi)/6 and the angle between sides B and C is pi/12π12pi/12. If side B has a length of 2, what is the area of the triangle?