Question Number 6890 by Tawakalitu. last updated on 01/Aug/16 Commented by Yozzii last updated on 02/Aug/16 $${Let}\:{EC}={x}.\:{BD}=\mathrm{3},\:{DA}=\mathrm{5} \\ $$$${BD}={BE}\:{since}\:{BD}\:{and}\:{BE}\:{are}\:{lines} \\ $$$${tangent}\:{to}\:{the}\:{same}\:{circle}\:{from}\: \\ $$$${the}\:{common}\:{point}\:{B}. \\ $$$$\therefore\:{BE}=\mathrm{3}.\:{Similarly},\:{DA}={AF}=\mathrm{5},…
Question Number 6872 by hanandmishra026@gmail.com last updated on 31/Jul/16 $${Eliminate}\:\theta\:\:{if}\:{x}={a}\:\mathrm{sec}^{\mathrm{3}} \theta\:{and}\:\:{y}={b}\:\mathrm{tan}^{\mathrm{3}} \theta \\ $$ Commented by Yozzii last updated on 31/Jul/16 $${sec}\theta=\left(\frac{{x}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} ,\:{tan}\theta=\left(\frac{{y}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${sec}^{\mathrm{2}}…
Question Number 6852 by Tawakalitu. last updated on 31/Jul/16 Commented by Tawakalitu. last updated on 31/Jul/16 $${Please}\:{i}\:{need}\:{solution}\:{here}….\:{to}\:{solve}\:{for}\:{x} \\ $$ Answered by Rasheed Soomro last updated…
Question Number 6806 by Tawakalitu. last updated on 27/Jul/16 Answered by Yozzii last updated on 28/Jul/16 $${We}\:{have}\:{that}\:\angle{QSR}=\mathrm{40}°.\:{Let}\:\omega\:{be}\: \\ $$$${circle}\:{PQRS}.\:{Points}\:{S}\:{and}\:{P}\:{lie}\:{on} \\ $$$$\omega\:{on}\:{the}\:{same}\:{side}\:{of}\:{line}\:{QR}.\: \\ $$$$\Rightarrow\angle{QSR}=\angle{QPR}\:\therefore\:\angle{QPR}=\mathrm{40}°. \\ $$$${TU}\:{is}\:{a}\:{straight}\:{line},\:{so}\:…
Question Number 137864 by n0y0n last updated on 07/Apr/21 $$\:\mathrm{if}\:, \\ $$$$\:\:\:\:\mathrm{acos}\alpha+\mathrm{bcos}\beta=\mathrm{Acos}\emptyset \\ $$$$\:\mathrm{proof}\:\mathrm{that}\: \\ $$$$\:\:\:\:\mathrm{asin}\alpha+\mathrm{bsin}\beta=\mathrm{Asin}\emptyset \\ $$ Commented by mr W last updated on…
Question Number 137861 by SLVR last updated on 07/Apr/21 Commented by SLVR last updated on 07/Apr/21 $${kindly}\:{help}\:{me}…{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 137851 by liberty last updated on 07/Apr/21 $${If}\:\mathrm{2}^{\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{3sin}\:\theta+\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} \theta+\mathrm{3sin}\:\theta} \:=\:\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:. \\ $$ Answered by bobhans last updated on 07/Apr/21…
Question Number 6750 by Tawakalitu. last updated on 22/Jul/16 Answered by Yozzii last updated on 22/Jul/16 $${Let}\:\angle{QPR}=\angle{RPS}=\alpha>\mathrm{0},\:\angle{PQR}=\beta>\mathrm{0}\:{and}\:\angle{PSR}=\delta>\mathrm{0}. \\ $$$${By}\:{the}\:{sine}\:{rule},\:{in}\:\bigtriangleup{QPR},\: \\ $$$$\frac{{QR}}{{sin}\alpha}=\frac{{QP}}{{sin}\angle{PRQ}}=\frac{\mathrm{3}{a}}{{sin}\left(\pi−\left(\alpha+\beta\right)\right)} \\ $$$$\Rightarrow{QR}=\frac{\mathrm{3}{asin}\alpha}{{sin}\left(\alpha+\beta\right)}. \\ $$$${In}\:\bigtriangleup{PRS},\:{by}\:{the}\:{sine}\:{rule}\:{we}\:{have}…
Question Number 137813 by bramlexs22 last updated on 07/Apr/21 $$\mathrm{cos}\:\left(\mathrm{arctan}\:\left(\frac{\mathrm{21}}{\mathrm{60}}\right)\right)=? \\ $$ Answered by mr W last updated on 07/Apr/21 $${say}\:{t}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{21}}{\mathrm{60}} \\ $$$$\mathrm{tan}\:{t}=\frac{\mathrm{21}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{20}} \\…
Question Number 137811 by bramlexs22 last updated on 07/Apr/21 $${Given}\:\begin{cases}{\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:{x}}\end{cases} \\ $$$${where}\:\mathrm{270}°\:<\:{y}<\:\mathrm{360}°\:.\:{find} \\ $$$$\mathrm{sin}\:\mathrm{2}{y}\:. \\ $$ Answered by EDWIN88 last updated on 07/Apr/21 $$\Leftrightarrow\:\mathrm{cos}\:\left({x}−{y}\right)+\mathrm{cos}\:\left({x}+{y}\right)=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{cos}\:{x} \\…