Question Number 92518 by I want to learn more last updated on 07/May/20 Answered by mr W last updated on 07/May/20 Commented by mr W…
Question Number 157965 by Tawa11 last updated on 30/Oct/21 Commented by Tawa11 last updated on 30/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{shaded}\:\mathrm{angles}. \\ $$ Answered by mr W last updated…
Question Number 26858 by $@ty@m last updated on 30/Dec/17 $${Prove}\:{that}\:{the}\:{internal}\:{bisector}\:{of}\:{an}\:{angle}\: \\ $$$${of}\:{a}\:{triangle}\:{divides}\:{the}\:{opposite} \\ $$$${sides}\:{in}\:{the}\:{ratio}\:{of}\:{the}\:{sides}\:{containing} \\ $$$${the}\:{angle}. \\ $$ Answered by mrW1 last updated on 30/Dec/17…
Question Number 26824 by kaivan.ahmadi last updated on 30/Dec/17 Answered by mrW1 last updated on 30/Dec/17 $$\angle{CBD}=\mathrm{60}−\mathrm{45}=\mathrm{15}° \\ $$$$\frac{{DB}}{\mathrm{sin}\:\angle{C}}=\frac{{CD}}{\mathrm{sin}\:\angle{CBD}} \\ $$$$\Rightarrow{DB}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\mathrm{15}}×{n}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){n} \\ $$$${AB}=\sqrt{{AD}^{\mathrm{2}} +{DB}^{\mathrm{2}} −\mathrm{2}×{AD}×{DB}×\mathrm{cos}\:\angle{ADB}}…
Question Number 26776 by ajfour last updated on 29/Dec/17 Commented by ajfour last updated on 29/Dec/17 $${Find}\:\mathrm{cot}\phi\:{in}\:{terms}\:{of}\:{c}\:. \\ $$ Commented by ajfour last updated on…
Question Number 26772 by Rasheed.Sindhi last updated on 29/Dec/17 $$\mathrm{With}\:\mathrm{a}\:\mathrm{center}\:\boldsymbol{\mathrm{on}}\:\mathrm{a}\:\mathrm{given}\:\mathrm{circle}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{r}\:,\mathrm{an}\:\mathrm{arc}\:\mathrm{has}\:\mathrm{been}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{order} \\ $$$$\mathrm{to}\:\mathrm{divide}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{two}\:\mathrm{equal} \\ $$$$\left(\mathrm{in}\:\mathrm{area}\right)\:\mathrm{parts}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\:\mathrm{r}\:\left(\mathrm{radius}\:\mathrm{of}\:\mathrm{given}\:\mathrm{circle}\right)?\: \\ $$ Answered by mrW1…
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Question Number 92269 by Power last updated on 05/May/20 Answered by mr W last updated on 05/May/20 $${let}\:{BD}=\mathrm{1} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\mathrm{51}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{25}+\mathrm{51}\right)} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{sin}\:\mathrm{51}}{\mathrm{sin}\:\mathrm{76}} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{99}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{15}+\mathrm{99}\right)} \\…
Question Number 26723 by Rasheed.Sindhi last updated on 28/Dec/17 $$\mathcal{T}{ake}\:{a}\:{point}\:\boldsymbol{{on}}\:{a}\:{given}\:{circle} \\ $$$${as}\:{a}\:{center}\:{and}\:{draw}\:{an}\:{arc} \\ $$$${which}\:{divide}\:{the}\:{given}\:{circle} \\ $$$${into}\:{two}\:{equal}\left(\mathrm{in}\:\mathrm{area}\right)\:{regions}.{Use}\:{only} \\ $$$${Eucledean}\:{tools}. \\ $$ Commented by prakash jain last…
Question Number 26711 by ktomboy1992 last updated on 28/Dec/17 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{tringle}.\mathrm{prove}\:\mathrm{it}? \\ $$ Commented by prakash jain last updated on 28/Dec/17 $$\mathrm{did}\:\mathrm{u}\:\mathrm{forget}\:\mathrm{to}\:\mathrm{add}\:\mathrm{a}\:\mathrm{picture}\:\mathrm{as}\:\mathrm{comment}? \\ $$ Terms of…