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Question Number 211251 by hardmath last updated on 01/Sep/24 Commented by a.lgnaoui last updated on 04/Sep/24 Commented by a.lgnaoui last updated on 05/Sep/24 $$\frac{\mathrm{sin}\:\mathrm{C}}{\mathrm{MN}}=\frac{\mathrm{sin}\:\mathrm{M}}{\mathrm{BC}−\mathrm{AB}}=\frac{\mathrm{sin}\:\mathrm{N}}{\mathrm{CD}−\mathrm{AD}}\left(\mathrm{1}\right) \…
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