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As shown in the figure, in △ ABC, ∠ ACB = 90 °, AC = BC, be ⊥ CE perpendicular foot is e, ad ⊥ CE perpendicular foot is D, ad = 2.5cm, be = 0.7cm, calculate the length of de
In △ CBE and △ ACD, if ∠ e = ∠ ADC = 90 °∠ CBE = ∠ acdbc = AC, ｜ △ CBE ≌ ACD (AAS), be = CD = 0.7cm, CE = ad = 2.5cm, then de = ce-cd = 2.5-0.7 = 1.8cm
In the triangle ABC, if AB = AC and Tanc = 3 / 2, then SINB =, COSC=
∵AB=AC
х B = C х both B and C are acute angles
∵tan C=3/2=sinC/cosC
Ψ sinc = 3cosc / 2 substituting (sinc) ^ 2 + (COSC) ^ 2 = 1
Sinc = 6 √ 13 / 12
cosC=4√13/12
That is SINB = 6 √ 13 / 12
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