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Question-6750

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}…

Let-f-x-x-2-2x-3-x-1-amp-g-x-1-x-4-x-4-then-the-number-of-real-solutions-of-equation-f-x-g-x-is-

Question Number 137817 by bramlexs22 last updated on 07/Apr/21 $${Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3};\:{x}\geqslant\mathrm{1}\:\& \\ $$$${g}\left({x}\right)=\mathrm{1}+\sqrt{{x}+\mathrm{4}}\:;\:{x}\geqslant−\mathrm{4}\:{then} \\ $$$${the}\:{number}\:{of}\:{real}\:{solutions} \\ $$$${of}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right)\:{is}\:… \\ $$ Answered by MJS_new last updated on…

cos-arctan-21-60-

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}} \\…

P-and-Q-are-partners-in-a-venture-P-contributed-20-000-for-nine-month-and-Q-contributed-50-000-for-one-year-find-each-person-share-of-profit-of-6-300-

Question Number 6743 by Tawakalitu. last updated on 20/Jul/16 $${P}\:{and}\:{Q}\:{are}\:{partners}\:{in}\:{a}\:{venture},\:\:{P}\:\:{contributed}\:#\mathrm{20},\mathrm{000}\:{for} \\ $$$${nine}\:{month},\:{and}\:{Q}\:{contributed}\:#\mathrm{50},\mathrm{000}\:{for}\:{one}\:{year}.\:{find}\:{each} \\ $$$${person}\:{share}\:{of}\:\:{profit}\:{of}\:#\mathrm{6},\mathrm{300} \\ $$$$ \\ $$ Answered by Rasheed Soomro last updated on…

Given-cos-x-y-1-1-2-cos-x-cos-x-y-1-1-3-cos-x-where-270-lt-y-lt-360-find-sin-2y-

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} \\…