Question Number 203232 by Mathstar last updated on 13/Jan/24 Answered by mr W last updated on 13/Jan/24 Commented by mr W last updated on 13/Jan/24…
Question Number 203195 by ajfour last updated on 12/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203187 by Amidip last updated on 12/Jan/24 Answered by mr W last updated on 12/Jan/24 $${supposed}:\:{AB}\bot{AC},\:{AD}\bot{BC} \\ $$$$\frac{\mathrm{45}}{{x}}=\frac{{x}+\mathrm{48}}{\mathrm{45}}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{45}^{\mathrm{2}} =\mathrm{0} \\…
Question Number 203211 by ajfour last updated on 12/Jan/24 Answered by MM42 last updated on 12/Jan/24 $${p}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cosa} \\ $$$${q}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{sina} \\ $$$$\Rightarrow\left({p}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\left({q}^{\mathrm{2}}…
Question Number 203115 by Mathstar last updated on 10/Jan/24 Commented by Mathstar last updated on 10/Jan/24 $$\mathrm{Send}\:\mathrm{an}\:\mathrm{explanatory}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you} \\ $$ Answered by mr W last updated…
Question Number 203107 by cherokeesay last updated on 10/Jan/24 Commented by cherokeesay last updated on 10/Jan/24 Find the length x. Answered by AST last updated…
Question Number 203062 by ajfour last updated on 09/Jan/24 Answered by ajfour last updated on 09/Jan/24 $$\mathrm{tan}\:\theta=\frac{{R}−\left(\mathrm{1}−{R}\right)}{{x}−{c}}=\frac{\mathrm{1}}{{x}} \\ $$$$\&\:\:{x}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\mathrm{1}−{R}\right)^{\mathrm{2}} =\mathrm{2}{R}−\mathrm{1} \\ $$$$\Rightarrow\:\frac{{x}−{c}}{{x}}=\frac{\mathrm{2}{R}−\mathrm{1}}{\mathrm{1}}={x}^{\mathrm{2}} \\…
Question Number 203055 by mr W last updated on 08/Jan/24 Commented by mr W last updated on 08/Jan/24 $${is}\:{it}\:{possible}\:{to}\:{determine}\:{the}\: \\ $$$${unknown}\:{area}\:{in}\:{terms}\:{of}\:{other} \\ $$$${partial}\:{areas}\:{A},{B},{C},{D},{E}? \\ $$…
Question Number 203018 by mr W last updated on 07/Jan/24 Commented by mr W last updated on 07/Jan/24 $$\left[{an}\:{unsolved}\:{old}\:{question}\:{Q}\mathrm{202864}\right] \\ $$ Commented by esmaeil last…
Question Number 203017 by mnjuly1970 last updated on 07/Jan/24 Answered by som(math1967) last updated on 07/Jan/24 $${let}\:{ar}\:{ofEDCF}={a}\:,\bigtriangleup{AEB}={b} \\ $$$$\frac{\bigtriangleup{ABD}}{\bigtriangleup{BCD}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{Green}+{b}}{{Magenta}+{a}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\bigtriangleup{ABF}}{\bigtriangleup{ACF}}=\frac{{x}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{Majenta}+{b}}{{Green}+{a}}=\frac{{x}}{\mathrm{6}}…