Question Number 125070 by Algoritm last updated on 08/Dec/20 Commented by mr W last updated on 08/Dec/20 $${x}\bot{y}\bot{z}\:{is}\:{not}\:{possible}. \\ $$$${do}\:{you}\:{mean}\:{x}\bot{a},\:{y}\bot{b},\:{z}\bot{c}? \\ $$$${but}\:{any}\:{way},\:{you}\:{can}\:{not}\:{determine} \\ $$$${the}\:{area}\:{of}\:{triangle}\:{with}\:{given} \\…
Question Number 190565 by cherokeesay last updated on 06/Apr/23 Answered by mr W last updated on 06/Apr/23 $$\frac{{CB}}{\mathrm{sin}\:{x}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{30}°\right)}{{CB}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{60}°\right)}{{DB}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{30}°\right)}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{60}°\right)}{\mathrm{sin}\:\mathrm{30}°} \\…
Question Number 125016 by Ggjj last updated on 07/Dec/20 $$\mathrm{6}+\mathrm{5}=\mathrm{11} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124953 by Algoritm last updated on 07/Dec/20 Answered by mr W last updated on 07/Dec/20 $${say}\:{AC}=\mathrm{1} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{50}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\mathrm{50}} \\ $$$${similarly} \\…
Question Number 124928 by bemath last updated on 07/Dec/20 $$\:{What}\:{are}\:{the}\:{exact}\:{values}\:{of}\:{k} \\ $$$${for}\:{which}\:{the}\:{line}\:{y}={kx}+\mathrm{3} \\ $$$${is}\:{tangent}\:{to}\:{the}\:{circle}\:{with} \\ $$$${centre}\:\left(\mathrm{6},\mathrm{3}\right)\:{and}\:{radius}\:\mathrm{2}? \\ $$ Answered by mr W last updated on…
Question Number 124869 by Algoritm last updated on 06/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59287 by pete last updated on 07/May/19 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2x}−\mathrm{4y}−\mathrm{5}=\mathrm{0}\:\mathrm{with}\:\mathrm{centr} \\ $$$$\mathrm{0}\:\mathrm{is}\:\mathrm{cut}\:\mathrm{by}\:\mathrm{a}\:\mathrm{line}\:\mathrm{y}=\mathrm{2x}+\mathrm{5}\:\mathrm{at}\:\mathrm{points}\:\mathrm{P}\:\mathrm{and}\:\mathrm{Q}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{QO}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{PO}. \\ $$ Answered by tanmay last updated on 07/May/19…
Question Number 190340 by ajfour last updated on 01/Apr/23 Commented by a.lgnaoui last updated on 01/Apr/23 Commented by a.lgnaoui last updated on 01/Apr/23 Commented by…
Question Number 190269 by normans last updated on 30/Mar/23 Commented by a.lgnaoui last updated on 02/Apr/23 Answered by a.lgnaoui last updated on 02/Apr/23 $${The}\:{Area}\:{of}\:\:\bigtriangleup{AEB}\:\:{ABC}\:{and}\:\bigtriangleup{AFC} \\…
Question Number 124722 by I want to learn more last updated on 05/Dec/20 Answered by mr W last updated on 05/Dec/20 Commented by mr W…