Category: Mensuration

Question Number 193161 by Mingma last updated on 05/Jun/23 Answered by a.lgnaoui last updated on 05/Jun/23 $$\mathrm{Triangle}\:\mathrm{ABC}\:\:\mathrm{equilaterale}\: \\ $$$$\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{NP}}=\boldsymbol{\mathrm{MP}} \\ $$$$\measuredangle\mathrm{BAH}=\mathrm{60}=\measuredangle\mathrm{DAH} \\ $$$$\Rightarrow\:\measuredangle\mathrm{ADH}=\mathrm{30}°\:\:\:\measuredangle\mathrm{AMN}=\mathrm{90}−\mathrm{30}=\mathrm{60} \\ $$$$\mathrm{MNP}\:\:\:\mathrm{Triangle}\:\mathrm{equilaterale}…

Question Number 64778 by aliesam last updated on 21/Jul/19 Commented by som(math1967) last updated on 21/Jul/19 $${join}\:{O},{Q}\:\:{O},{N}\:{O},{M}\:\:{M},{Q}\:{N},{Q} \\ $$$${now}\:{ON}={NQ}={OQ}={OM}={MQ} \\ $$$$\therefore\angle{NOM}=\mathrm{60}+\mathrm{60}=\mathrm{120}° \\ $$$$\angle{MLN}=\frac{\mathrm{1}}{\mathrm{2}}\angle{NOM}=\mathrm{60}° \\ $$$$\angle{NKM}=\angle{NOM}=\mathrm{120}°\:\:\left[{subtend}\:{on}\:{same}\:{segment}\right]…

Question Number 193069 by Mingma last updated on 03/Jun/23 Answered by ajfour last updated on 03/Jun/23 Commented by ajfour last updated on 03/Jun/23 $${a}=\mathrm{17cos}\:\theta \\…

Question Number 192966 by Mingma last updated on 01/Jun/23 Answered by aleks041103 last updated on 01/Jun/23 $${at}\:{the}\:{intersection}\:\left(\mathrm{0},\mathrm{0}\right)\in\mathbb{R}^{\mathrm{2}} . \\ $$$${the}\:{whole}\:{circle}\:{is} \\ $$$$\left({x}−\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{125}}{\mathrm{16}}=\left(\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} \\…

Question Number 127382 by I want to learn more last updated on 29/Dec/20 Answered by mr W last updated on 29/Dec/20 Commented by mr W…