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Category: Trigonometry

Question-49660

Question Number 49660 by Rio Michael last updated on 08/Dec/18 Commented by Rio Michael last updated on 08/Dec/18 $${Above}\:{is}\:{a}\:{Circle}\:{with}\:{centre}\:{O}\:.{If}\:{it}\:{has}\:{radius}\:{of}\:\mathrm{10}{cm} \\ $$$${find}\:{the}\:{lenght}\:{of}\:{the}\:{line}\:{AC}.{Note}\::\:\angle{AOC}=\mathrm{30}° \\ $$ Answered by…

Question-49605

Question Number 49605 by Rio Michael last updated on 08/Dec/18 Commented by Rio Michael last updated on 08/Dec/18 $${The}\:{figure}\:{above}\:{has}\:\bigtriangleup{ABC}\:{and}\:\bigtriangleup{MBD}\:{where}\:{AC}\:{is}\:{parallel} \\ $$$${to}\:{DM}.{Find}\:{the}\:{lenght}\:{AM}\:{if} \\ $$$$\left.{a}\right)\:{BM}\:=\:\mathrm{6}.\mathrm{5}{m} \\ $$$$\left.{b}\right)\:{Given}\:{the}\:{the}\:{area}\:{of}\:\bigtriangleup{ABC}\:=\:\mathrm{120}{m}^{\mathrm{2}}…

solve-tan-x-cot-x-p-sec-x-cos-x-q-

Question Number 115121 by bemath last updated on 23/Sep/20 $${solve}\:\begin{cases}{\mathrm{tan}\:{x}\:+\:\mathrm{cot}\:{x}\:=\:{p}}\\{\mathrm{sec}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{q}}\end{cases} \\ $$ Commented by ATHISHHUZAIN last updated on 23/Sep/20 $${f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}{x}+\mathrm{1}\:\:\:\:\:{o}}\\{\mathrm{sec}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{q}}\end{cases} \\ $$ Commented by bemath…

Q-1-Find-the-domain-of-f-x-cos-1-x-4-x-2tan-1-x-sin-lnx-3x-2-7-a-sin-x-3cos-x-ln-cos-1-x-2-where-x-reprents-the-fractionare-pa

Question Number 115022 by dw last updated on 23/Sep/20 $$\left[\boldsymbol{{Q}}.\mathrm{1}\:\right]\:\:\:{Find}\:{the}\:{domain}\:{of} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\frac{\lfloor{cos}^{−\mathrm{1}} \left({x}^{\mathrm{4}} \right)\rfloor+\mid\lfloor{x}−\mathrm{2}{tan}^{−\mathrm{1}} \left({x}\right)\rfloor\mid+\sqrt{{sin}\left({lnx}\right)}}{\left\{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{7}\right\}+{a}^{\sqrt{{sin}\left({x}\right)+\mathrm{3}{cos}\left({x}\right)}} +{ln}\:{cos}\left(\frac{\mathrm{1}}{\:\sqrt{−{x}^{\mathrm{2}} }}\right)} \\ $$$${where}\:\left\{{x}\right\}\:{reprents}\:\:{the}\:{fractionare}\:{part}\:{of}\:{x}: \\ $$$$ \\…