Question Number 199339 by necx122 last updated on 01/Nov/23 Commented by necx122 last updated on 01/Nov/23 $${find}\:\angle{QSR}\:{where}\:{O}\:{is}\:{the}\:{centre} \\ $$$$\angle{OTQ}\:=\mathrm{15},\:\angle{TOR}=\mathrm{110} \\ $$ Commented by AST last…
Question Number 199286 by ajfour last updated on 31/Oct/23 Answered by ajfour last updated on 31/Oct/23 $${Let}\:{height}\:{of}\://{gm}={h} \\ $$$${AB}=\mathrm{1}\:\: \\ $$$$\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }={p} \\ $$$$\frac{{r}}{\mathrm{2}}\left(\mathrm{1}+{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\right)=\frac{{p}}{\mathrm{2}}…
Question Number 199242 by Tawa11 last updated on 30/Oct/23 Answered by mr W last updated on 30/Oct/23 Commented by mr W last updated on 30/Oct/23…
Question Number 199234 by mr W last updated on 30/Oct/23 Commented by mr W last updated on 30/Oct/23 $${the}\:{side}\:{length}\:{of}\:{the}\:{squares}\:{is}\:\mathrm{1}. \\ $$$${find}\:{the}\:{area}\:{of}\:{the}\:{isosceles}\:{right} \\ $$$${angled}\:{triangle}. \\ $$…
Question Number 199149 by ajfour last updated on 28/Oct/23 Commented by ajfour last updated on 28/Oct/23 $${No};\:\:{rather}\:{a}=\mathrm{1}<{b}.\:\:{Find}\:{R}={f}\left({b}\right). \\ $$ Answered by ajfour last updated on…
Question Number 199089 by mr W last updated on 28/Oct/23 Answered by ajfour last updated on 28/Oct/23 Commented by ajfour last updated on 28/Oct/23 $${OB}^{\:\mathrm{2}}…
Question Number 199046 by mr W last updated on 27/Oct/23 Commented by mr W last updated on 27/Oct/23 $${as}\:{Q}\mathrm{198763},\:{but}\:{the}\:{balls}\:{are}\:{not}\:{on} \\ $$$${a}\:{table},\:{but}\:{inside}\:{a}\:{large}\:{spherical} \\ $$$${bowl}\:{with}\:{radius}\:{R}. \\ $$…
Question Number 199023 by mr W last updated on 26/Oct/23 Commented by mr W last updated on 26/Oct/23 $${if}\:{AX}={YB},\:{prove}\:\angle{ADX}=\angle{YDC}. \\ $$ Answered by mr W…
Question Number 199006 by adhigenz last updated on 26/Oct/23 $$\mathrm{Given}\:\mathrm{that}\:{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{trapezium}\:\mathrm{such}\:\mathrm{that}\:{AD}//{BC}. \\ $$$$\mathrm{The}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABD}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{BCD}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABC}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{ADC}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198933 by cherokeesay last updated on 25/Oct/23 Commented by mr W last updated on 26/Oct/23 Commented by Rasheed.Sindhi last updated on 26/Oct/23 $$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{diagram}}\:\boldsymbol{\mathrm{sir}}!…