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Category: Geometry

Question-209932

Question Number 209932 by cherokeesay last updated on 26/Jul/24 Answered by mahdipoor last updated on 26/Jul/24 $${BD}^{\mathrm{2}} =\mathrm{100}+\mathrm{49}−\mathrm{140}{cosC}=\mathrm{121}+\mathrm{64}−\mathrm{176}{cosA} \\ $$$$\frac{{BD}}{{sinC}}=\frac{{BD}}{{sinA}}=\mathrm{2}{R}\:\Rightarrow{C}+{A}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{149}−\mathrm{140}{cosC}=\mathrm{185}−\mathrm{176}{cosA} \\ $$$$\Rightarrow{cosC}=\frac{\mathrm{185}−\mathrm{149}}{−\mathrm{176}−\mathrm{140}}=−\frac{\mathrm{9}}{\mathrm{79}} \\…

Question-209889

Question Number 209889 by Tawa11 last updated on 24/Jul/24 Answered by som(math1967) last updated on 25/Jul/24 $${let}\:{PQ}={QS}={QR}={x} \\ $$$${cos}\angle{PQS}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{72}}{{x}^{\mathrm{2}} }…

Question-209813

Question Number 209813 by mr W last updated on 22/Jul/24 Answered by mahdipoor last updated on 22/Jul/24 $${R}=\:{radius}\:{red}\:{circle} \\ $$$${r}\:=\:{radius}\:{white}\:{circle} \\ $$$$\begin{cases}{\mathrm{2}{R}=\mathrm{8}+\mathrm{2}+\mathrm{2}{r}}\\{{C}_{\mathrm{1}} {C}_{\mathrm{2}} =\sqrt{{r}^{\mathrm{2}} +\left({R}−\left({r}+\mathrm{2}\right)\right)^{\mathrm{2}}…