Question Number 146878 by gsk2684 last updated on 16/Jul/21 $${in}\:\Delta{ABC}\:{if}\:\mathrm{sin}\:^{\mathrm{2}} {A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}+ \\ $$$$\mathrm{cos}\:{B}\mathrm{cos}\:{C}=\mathrm{1}\:{then}\:{the}\:{triangle}\:{is} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 146876 by gsk2684 last updated on 16/Jul/21 $${if}\:{the}\:{maximum}\:{value}\:{of}\: \\ $$$$\mathrm{4sin}\:^{\mathrm{2}} {x}+\mathrm{3cos}\:^{\mathrm{2}} {x}+\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\mathrm{3} \\ $$$${is}\:{a}+\sqrt{{b}}\:{then}\:{find}\:{a}+{b} \\ $$ Answered by liberty last updated on 16/Jul/21…
Question Number 146875 by gsk2684 last updated on 16/Jul/21 $${In}\:{a}\:{triangle}\:{ABC},\:{if}\: \\ $$$$\frac{\mathrm{sin}\:{A}}{\mathrm{5}−{x}}=\frac{\mathrm{sin}\:{B}}{\mathrm{3}{x}−\mathrm{1}}=\frac{\mathrm{sin}\:{C}}{\mathrm{2}{x}+\mathrm{5}}\:{then}\:{find} \\ $$$$\:{integral}\:{solutions}\:{x}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 146874 by gsk2684 last updated on 16/Jul/21 $${let}\:{the}\:{line}\:{joining}\:{through}\: \\ $$$${orthocenter}\:{and}\:{circumcenter}\: \\ $$$${of}\:{a}\:{triangle}\:{ABC}\:{is}\:{parallel}\:{to}\: \\ $$$${the}\:{base}\:{BC}\:{then}\:{find}\:\:\mathrm{tan}\:{B}.\mathrm{tan}\:{C} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 146868 by gsk2684 last updated on 16/Jul/21 $${if}\:\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{sec}\:{x}={a}+\mathrm{1}\:{has}\:{at}\:{least}\: \\ $$$${one}\:{solution}\:{then}\:{find}\:{the}\:{complete}\:{set} \\ $$$${of}\:{values}\:{of}\:\:'{a}'? \\ $$ Answered by Olaf_Thorendsen last updated on 16/Jul/21 $$\mathrm{tan}^{\mathrm{2}}…
Question Number 15786 by tawa tawa last updated on 13/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{7}}\right)\:−\:\mathrm{cos}\left(\frac{\mathrm{2x}}{\mathrm{7}}\right)\:+\:\mathrm{cos}\left(\frac{\mathrm{3x}}{\mathrm{7}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by tawa tawa last updated on 14/Jun/17 $$\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{this}. \\…
Question Number 146850 by Apor_mu_calculus last updated on 16/Jul/21 $$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 15759 by prakash jain last updated on 13/Jun/17 $$\mathrm{Let}\:\mathrm{us}\:\mathrm{call}\:\mathrm{complex}\:\mathrm{triangle}\:\mathrm{which} \\ $$$$\mathrm{has}\:\mathrm{either}\:\mathrm{sides}\:\mathrm{or}\:\mathrm{angles}\:\mathrm{are} \\ $$$$\mathrm{complex}\:\mathrm{numbers}. \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\in\mathbb{R}\:\mathrm{which}\:\mathrm{are}\:\mathrm{sides}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{complex}\:\mathrm{triangle}\:\mathrm{which}\:\mathrm{need} \\ $$$$\mathrm{not}\:\mathrm{satisfy}\:\mathrm{triangle}\:\mathrm{inequality}. \\ $$$$\mathrm{say}\:{a}=\mathrm{1},{b}=\mathrm{2}\:\mathrm{and}\:{c}=\mathrm{4}. \\ $$$$\mathrm{Prove}\:\left(\mathrm{or}\:\mathrm{counter}\:\mathrm{example}\right)…
Question Number 15760 by Tinkutara last updated on 13/Jun/17 $$\mathrm{If}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{are}\:{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}, \\ $$$$\mathrm{2}{x}\:+\:\mathrm{1}\:\mathrm{and}\:{x}^{\mathrm{2}} \:−\:\mathrm{1},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{greatest} \\ $$$$\mathrm{angle}\:\mathrm{is}\:\mathrm{120}°.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{triangle}\:\mathrm{exist}. \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated…
Question Number 15761 by Tinkutara last updated on 13/Jun/17 $$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\frac{\mathrm{2}\:\mathrm{cos}\:{A}}{{a}}\:+\:\frac{\mathrm{cos}\:{B}}{{b}}\:+\:\frac{\mathrm{2}\:\mathrm{cos}\:{C}}{{c}} \\ $$$$=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ac}}\:,\:\mathrm{prove}\:\mathrm{that}\:\angle{A}\:=\:\mathrm{90}°. \\ $$ Answered by ajfour last updated on 13/Jun/17 $$\frac{\mathrm{2cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\frac{\mathrm{2cos}\:{C}}{{c}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}} \\…