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Category: Geometry

Question-224337

Question Number 224337 by wongb1506 last updated on 04/Sep/25 Answered by BaliramKumar last updated on 04/Sep/25 $${cos}\left({A}=\mathrm{60}°\right)\:=\:\frac{\mathrm{5}^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\left({x}−\mathrm{6}\right)} \\ $$$${x}=\mathrm{14}\:\checkmark\:\:\:\:\:{or}\:\:\:\:\:\:\cancel{\mathrm{3}}\: \\ $$$$ \\…

Question-224336

Question Number 224336 by wongb1506 last updated on 04/Sep/25 Answered by som(math1967) last updated on 04/Sep/25 $${let}\:{EC}={x},\:\angle{BEC}=\alpha \\ $$$$\therefore\angle{AED}=\mathrm{180}−\mathrm{60}−\alpha=\mathrm{120}−\alpha \\ $$$$\:\frac{\mathrm{4}}{{x}}={sin}\alpha\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{4}} \\ $$$$\:\:\frac{\mathrm{3}}{{x}}={sin}\left(\mathrm{120}−\alpha\right)\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}} \\ $$$$\therefore\:\frac{{sin}\alpha}{\mathrm{4}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}}…

Question-224108

Question Number 224108 by MirHasibulHossain last updated on 21/Aug/25 Answered by mr W last updated on 21/Aug/25 $${i}\:{think}\:{if}\:{the}\:{three}\:{big}\:{blue}\:{circles}\: \\ $$$${have}\:{different}\:{radii}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\: \\ $$$${generally}\:{we}\:{have}…

Question-224013

Question Number 224013 by efronzo1 last updated on 14/Aug/25 Commented by efronzo1 last updated on 14/Aug/25 $$\mathrm{Given}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{diameter}\:,\:\angle\mathrm{AOD}=\mathrm{22}° \\ $$$$\:\mathrm{and}\:\angle\mathrm{BCD}=\:\mathrm{p}°. \\ $$$$\:\mathrm{Find}\:\mathrm{p}. \\ $$ Answered by…