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In the given figure, AB is the diameter of the circle, centered at O. If angleCOA = 60^(@), AB = 2r, AC = d, and CD = l, then l is equal to |
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Answer» `d sqrt3` `AC = d, OA = OB r, CD = BD = I, angleCOA = (pi)/(3)` CLEARLY `DeltaAOC` is equilateral `:. d = r` Also, `angleBOD = angleCOD = (2pi)/(3xx2) = (pi)/(3)` or `TAN.(pi)/(3) = (BD)/(OB) = (l)/(r) RARR l = r sqrt3 = d sqrt3` |
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