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