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Category: Geometry

Question-193293

Question Number 193293 by TUN last updated on 09/Jun/23 Answered by AST last updated on 09/Jun/23 $${h}={a}+{b}+{c} \\ $$$$\mid{AH}\mid^{\mathrm{2}} =\mid{b}+{c}\mid^{\mathrm{2}} =\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}=\left({b}+{c}\right)\left(\overset{−} {{b}}+\overset{−} {{c}}\right) \\…

Question-193239

Question Number 193239 by Mingma last updated on 08/Jun/23 Answered by a.lgnaoui last updated on 08/Jun/23 $$\measuredangle\mathrm{B}=\mathrm{180}−\mathrm{110}=\mathrm{70} \\ $$$$\bigtriangleup\mathrm{ABE}\:\:\:\:\measuredangle\mathrm{AEB}=\mathrm{180}−\mathrm{120}=\mathrm{60} \\ $$$$\bigtriangleup\mathrm{ACE}\:\:\:\:\measuredangle\mathrm{AEC}=\mathrm{180}−\mathrm{60}=\mathrm{120} \\ $$$$ \\ $$$$\bigtriangleup\mathrm{ABC}\:\:\:\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{AB}}=\frac{\mathrm{sin}\:\mathrm{70}}{\mathrm{BC}}…

Question-65380

Question Number 65380 by ajfour last updated on 29/Jul/19 Commented by ajfour last updated on 29/Jul/19 $${Find}\:{radius}\:{r}\:{of}\:{a}\:{circle}\:{whose} \\ $$$${center}\:{is}\:{on}\:{the}\:{circumference} \\ $$$${of}\:{the}\:{unit}\:{circle}\:{and}\:{whose} \\ $$$${arc}\:{length}\:{within}\:{the}\:{shown} \\ $$$${unit}\:{radius}\:{circle}\:{is}\:{a}\:{maximum}.…

Question-65366

Question Number 65366 by Tawa1 last updated on 29/Jul/19 Commented by Prithwish sen last updated on 29/Jul/19 $$\mathrm{the}\:\mathrm{length}\:\mathrm{is} \\ $$$$=\:\mathrm{3}\:+\:\mathrm{2}×\mathrm{2}\:+\mathrm{2}×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }\:+\:\mathrm{2}×\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{3}} }\:+\:…….…

Question-65334

Question Number 65334 by Tawa1 last updated on 28/Jul/19 Answered by mr W last updated on 28/Jul/19 $$\angle{D}=\mathrm{360}−\mathrm{126}−\mathrm{42}×\mathrm{2}=\mathrm{150}° \\ $$$$\frac{{DC}}{\mathrm{sin}\:{x}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{150}} \\ $$$$\Rightarrow{DC}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{150}}×{AC} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{126}}=\frac{{AB}}{\mathrm{sin}\:\left[\mathrm{42}−\left(\mathrm{30}−{x}\right)\right]}=\frac{{AB}}{\mathrm{sin}\:\left({x}+\mathrm{12}\right)} \\…