In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If SINB = 2sinc and A2 − B2 = 32bc, then a=______ ．

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If SINB = 2sinc and A2 − B2 = 32bc, then a=______ ．

∵ in △ ABC, the opposite sides of angles a, B and C are a, B, C, ∵ bsinb = csinc, ∵ SINB = bcsinc, ∵ SINB = 2sinc, ∵ BC = 2, that is, B = 2c, ∵ A2 − B2 = 32bc, ∵ a2-4c2 = 3c2, ∵ a = 7C, ∵ cosa = B2 + C2 − a22bc = 4C2 + C2 − 7c22 × 2C ×× C = - 12, ∵ a = 23 π respectively

In the triangle ABC, if a = 8, C = 14 and 5sina = 2sinc + SINB, find B

Suppose the radius of circumscribed circle is r, then a = 2rsina, B = 2rsinb, C = 2rsinc,
The two ends of 5sina = 2sinc + SINB are multiplied by 2R to get 5A = 2C + B,
So B = 5 * 8-2 * 14 = 12

In △ ABC, if Sina = 2sinbcosc, sin2a = sin2b + sin2c, then the shape of △ ABC ()
A. Isosceles right triangle B. isosceles triangle C. right triangle D. equilateral triangle

In this paper, we use sine theorem to simplify sin2a = sin2b + sin2c, and get: A2 = B2 + C2, ∵ ABC is right triangle, ∵ Sina = sin (B + C) = sinbcosc + cosbsinc = 2sinbcosc, ∵ sinccosb coscsinb = sin (C-B) = 0, ∵ C-B = 0, that is, B = C, then the shape of △ ABC is isosceles right triangle