Let P = (a + C, b) and q = (B − a, C − a). If P ‖ Q, then the size of angle c is______ ．

Let P = (a + C, b) and q = (B − a, C − a). If P ‖ Q, then the size of angle c is______ ．

Because P ∥ Q, a + CB − a = BC − a, B2 AB = c2-a2, that is, A2 + b2-c2 = ab. by cosine theorem COSC = A2 + B2 − c22ab = 12, C = π 3, so the answer is: π 3

Let m = (B + A, c) and N = (B-A, C-B). If M is perpendicular to N, then the value range of SINB + sinc is

M is perpendicular to N, so (B + a) (B-A) + C * (C-B) = 0A ^ 2 = B ^ 2 + C ^ 2-bca ^ 2 = B ^ 2 + C ^ 2-2bccos60 according to the cosine theorem, a = 60 degree, B + C = 120 degree, SINB + sinc = 2Sin (B + C) / 2 * cos (B-C) / 2 = 2 * √ 3 / 2cos (B-C) / 2 = √ 3cos (B-C) / 2 = √ 3cos (120-2c) / 2 = √ 3cos (60-c) C takes 0 to 120

As shown in the figure, in isosceles △ ABC, ab = AC, ad ⊥ BC at point D, CF ∥ AB, P is a point on ad, connecting and extending BP, crossing AC at point E, crossing CF at point F, then
, BP ^ 2 = PE * pf?

So what?! @!

As shown in the figure, in △ ABC, ad is the bisector of ∠ BAC, BP ⊥ ad, and the perpendicular foot is p. it is known that ab = 5, BP = 2, AC = 9

It is proved that: to extend BP, to make AC in E, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\∠ ABC = ∠ Abe + ∠ EBC =3∠C．