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It involves solving triangles and vectors In △ ABC, the opposite sides of ∠ a, B, C are a, B, C, vector M = (a ^ 2, B ^ 2), vector n = (Tana, tanb), and vector M / / vector n. then what is the shape of △ ABC?
Vector M = (a ^ 2, B ^ 2), vector n = (Tana, tanb), ∵ vector M / / vector n ∵ A & # 178; tanb = B & # 178; Tana according to sine theorem a = 2rsina, B = 2rsinb ∵ Sin & # 178; a * SINB / CoSb = Sin & # 178; b * Sina / Cosa ∵ Sina / CoSb = SINB / Cosa ∵ sinacosa = sinbcosb ∵ sin2a = sin2
As shown in the figure, ∠ a + B + C + D + e equals ()
A. 180°B. 360°C. 540°D. 720°
∫∠ 1 is the outer angle of △ CEF, ∫∠ 1 = ∠ C + ∠ E; ∫∠ 2 is the outer angle of △ BDG, ∫ 2 = ∠ B + ∠ D, ∫ a + ∠ 1 + ∠ 2 = 180 ° and ∫ a + ∠ B + ∠ C + ∠ D + ∠ e = 180 °. Therefore, a
3ab^(-2) b^3)^(-2)/[a^(-1)b^(-2)
3ab^(-2) b^3)^(-2)/[a^(-1)b^(-2) c]^2
=3ab^(-2) b^(-6)/[a^(-2)b^(-4) c^2]
=3ab^(-8) *a^2b^4 c^(-2)
=3a^3b^(-4)c^(-2)
If 2 / A-2 / b = 1 / (a-b), then B / A + A / b =?
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