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• Pretty Cure Q1-5@1@
• 世界各國家及城市英文縮寫一問-----------10點-------------@1@
• Why the dipole moment of HF is greater than that of HI--

 

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Draw a line joining AB, and also joining OA and OB, where O is the centre of the circle. Now, as AR and BR are the tangents to the circle, so AR = BR (tangent property) Therefore, ∠RAB = ∠RBA = (180* - 64*)/2 = 58* Consider the quadrilateral OARB, As OA = OB = radius of the circle, so, ∠OAB = ∠OBA Let x be ∠AOB So, ∠OAB = ∠OBA = (180* - x)/2 = 90* - x/2 But note that AR is a tangent and OA is a radius, so, ∠OAR = 90* (line joining centre to tangent perp. tangent) So, 58* + (90* - x/2) = 90* x = 116* ∠AOB = 2∠ACB (∠ at centre twice ∠ at circumference) Therefore, ∠ACB = 116*/2 = 58* 4.a. AQ = PQ - PA = (11 - x) cm PB = PA = x cm (tangent property) BR = RP - PB = (9 - x) cm b. CR = BR = (9 - x)cm (tangent property) QC = AQ = (11 - x) cm (tangent property) As QR = QC + CR So, 10 = (11 - x) + (9 - x) 2x = 10 x = 5

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