In the triangle ABC, ∠ BCA = 90 ° and CD ⊥ AB is at point D. given AC = 3cm, BC = 4cm and ab = 5cm, what is the distance between point C and ab? RT The first answer is the best

Man, you're not good at math
SinB=AC/AB=CD/BC
So CD = AC * BC / AB = 3 * 4 / 5 = 12 / 5

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