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Category: Geometry

Let-ABC-be-a-triangle-with-AB-AC-and-BAC-30-Let-A-be-the-reflection-of-A-in-the-line-BC-B-be-the-reflection-of-B-in-the-line-CA-C-be-the-reflection-of-C-in-the-line-AB-Show-that-A-B-C

Question Number 23592 by Tinkutara last updated on 02/Nov/17 $$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:{AB}\:=\:{AC} \\ $$$$\mathrm{and}\:\angle{BAC}\:=\:\mathrm{30}°.\:\mathrm{Let}\:{A}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{A}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{BC};\:{B}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA};\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Show}\:\mathrm{that}\:{A}',\:{B}',\:{C}' \\ $$$$\mathrm{form}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$ Answered…

Question-89097

Question Number 89097 by TawaTawa1 last updated on 15/Apr/20 Commented by mr W last updated on 15/Apr/20 $$\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{10}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{10}} \\ $$$${s}=\mathrm{5}\sqrt{\mathrm{10}}−\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{10}}}\left(\mathrm{15}+\mathrm{5}\right)=\mathrm{3}\sqrt{\mathrm{10}} \\ $$$${area}\:={s}^{\mathrm{2}} =\mathrm{9}×\mathrm{10}=\mathrm{90}…

tan-6-x-dx-

Question Number 23539 by tapan das last updated on 01/Nov/17 $$\int\mathrm{tan}\:^{\mathrm{6}} \mathrm{x}\:\mathrm{dx} \\ $$ Answered by $@ty@m last updated on 01/Nov/17 $$\int\mathrm{tan}\:^{\mathrm{4}} {x}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\…

sec-2-x-x-dx-

Question Number 23418 by tapan das last updated on 30/Oct/17 $$\int\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{x}}\:/\sqrt{\mathrm{x}}\:\mathrm{dx} \\ $$ Answered by $@ty@m last updated on 30/Oct/17 $${Assume}\:\sqrt{{x}}={t}\:\&\:{proceed}. \\ $$$${Ans}\:\mathrm{2tan}\:\sqrt{{x}}+{C} \\…

Question-23408

Question Number 23408 by selestian last updated on 29/Oct/17 Answered by mrW1 last updated on 31/Oct/17 $$\mathrm{FE}=\frac{\mathrm{5}}{\mathrm{10}}×\mathrm{18}=\mathrm{9} \\ $$$$\mathrm{AC}=\frac{\mathrm{15}}{\mathrm{9}}×\left(\mathrm{5}+\mathrm{10}\right)=\mathrm{25} \\ $$$$\Rightarrow\mathrm{Answer}\:\left(\mathrm{b}\right) \\ $$ Terms of…