A triangle has sides A, B, and C. If the angle between sides A and B is $(pi)/12$ , the angle between sides B and C is $(3pi)/4$ , and the length of side B is 9, what is the area of the triangle?

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Answer 1 · 2016-06-27T09:21:03

Area of triangle is 14.777

Length of side b = 9

Angle between sides a and b = $pi/12$ = $/_C$

Angle between sides b and c = $(3pi)/4$ = $/_A$

Sum of the angles of a $triangle$ = $pi$

Hence, angle between c and a = $/_B$ = $pi - (pi/12 + (9pi)/12)$ = $((2pi)/12)$ = $pi/6$

Using sine rule = $a/sin A = b/ sin B = c/ sin C$ ,

we get $b/ sin (pi/6) =a/sin ((3pi)/4) = c/(sin (pi/12))$

$9/ (1/2) =a/sin ((3pi)/4) = c/(sin (pi/12))$

$a = 18* sin ((3pi)/4)$ = $18*(sqrt 2) / 2$ = $18 * 0.707106781$ = $12.7279$

A = $1/2*a*b*Sin C$ = $1/2*12.7279*9*Sin (pi/12)$

= $1/2*12.7279*9*2.58$ = 14.777

A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12(pi)/12, the angle between sides B and C is (3pi)/4(3pi)/4, and the length of side B is 9, what is the area of the triangle?