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Category: Geometry

Question-218475

Question Number 218475 by lmcp1203 last updated on 10/Apr/25 Answered by Nicholas666 last updated on 10/Apr/25 $$\bullet\angle{CBD}\:+\:\angle{BDC}\:+\:\angle{BCD}\:=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}+\mathrm{140}°+\mathrm{30}°=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}+\mathrm{170}°=\mathrm{180}° \\ $$$$\bullet\boldsymbol{{x}}=\mathrm{180}°−\mathrm{170}° \\ $$$$\bullet\boldsymbol{{x}}=\mathrm{10}°…

Question-218456

Question Number 218456 by Spillover last updated on 10/Apr/25 Answered by vnm last updated on 10/Apr/25 $$\mathrm{d}\:\mathrm{is}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{centers} \\ $$$$\varphi\:\mathrm{is}\:\mathscr{L}\cancel{\underbrace{ }} \\ $$ Commented by Spillover…

Question-218410

Question Number 218410 by Spillover last updated on 09/Apr/25 Answered by Nicholas666 last updated on 09/Apr/25 $$\:{given};{AB},{CD}\:{tangents}\:{to}\:{circle}\:{O}: \\ $$$${M}\:{midpoint}\:{of}\:{BC};\:{AD}\mid\mid{BC};\:{BD}={x},{AB}={y},{CD}={z} \\ $$$${prove}\:{x}^{\mathrm{2}} ={y}.{z} \\ $$$${solution} \\…