Question Number 194410 by necx122 last updated on 05/Jul/23
Commented by necx122 last updated on 05/Jul/23
$${I}'{ve}\:{dedicated}\:{sometime}\:{to}\:{solving}\:{the} \\ $$$${question}\:{but}\:{it}\:{seems}\:{as}\:{though}\:{some} \\ $$$${parameters}\:{are}\:{missing}.{Please}\:{help} \\ $$$${me}\:{out}.\: \\ $$
Answered by AST last updated on 05/Jul/23
$${The}\:{rectangle}\:{seems}\:{to}\:{be}\:{a}\:{square}.{Suppose}\:{it}\:{is} \\ $$$${Let}\:{x}\:{be}\:{the}\:{length}\:{of}\:{the}\:{side}\:{of}\:{the}\:{square} \\ $$$$\frac{{sin}\:\mathrm{90}°}{\mathrm{6}}=\frac{{sin}\:\mathrm{75}°}{{x}}\Rightarrow{x}=\mathrm{6}{sin}\mathrm{75}°=\frac{\mathrm{3}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}{cm} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{9}\left(\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}\right)}{\mathrm{4}}=\mathrm{9}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{18}+\mathrm{9}\sqrt{\mathrm{3}}{cm}^{\mathrm{2}} \\ $$$${Area}\:{of}\:{shaded}\:{area}=\frac{\mathrm{36}\sqrt{\mathrm{3}}}{\mathrm{4}}=\mathrm{9}\sqrt{\mathrm{3}}{cm}^{\mathrm{2}} \\ $$$$\Rightarrow{Area}\:{of}\:{unshaded}\:{area}=\mathrm{18}{cm}^{\mathrm{2}} \\ $$
Commented by necx122 last updated on 05/Jul/23
$${I}\:{really}\:{appreciate}\:{your}\:{steps}\:{sir}/{ma} \\ $$$${but}\:{the}\:{textbook}\:{says}\:{that}\:{the}\:{area}\:{is} \\ $$$$\mathrm{15}.\mathrm{5} \\ $$
Commented by AST last updated on 05/Jul/23
$${Does}\:{the}\:{textbook}\:{have}\:{a}\:{solution}? \\ $$
Commented by necx122 last updated on 05/Jul/23
$${not}\:{at}\:{all}.\:{There}'{s}\:{no}\:{solution}\:{just}\:{answers} \\ $$$${at}\:{the}\:{back}\:{of}\:{the}\:{book}. \\ $$
Commented by AST last updated on 05/Jul/23
$${Not}\:{always}\:{correct}. \\ $$
Commented by necx122 last updated on 06/Jul/23
$${okay}\:{sir}.\:{But}\:{please},\:{how}\:{did}\:{you}\:{get}\:{the} \\ $$$$\mathrm{75}°\:{you}\:{used}.\:{Can}\:{you}\:{maybe}\:{help}\:{me}\: \\ $$$${with}\:{a}\:{diagram}\:{that}\:{explains}\:{it}\:{properly}. \\ $$$${Thanks}\:{kn}\:{anticipation}. \\ $$$$ \\ $$
Commented by AST last updated on 06/Jul/23
$${Consider}\:\:\:\:\:\:\:\:{A}\:\:\:\:{X}\:\:\:\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Y}\:\:\:\:\:\:\:\:\:\:\:{B} \\ $$$${cosZ}\overset{\wedge} {{Y}B}={cosA}\overset{\wedge} {{Y}X}=\frac{{x}}{\mathrm{6}};{and}\:{since}\:{Z}\overset{\wedge} {{Y}B}\:{and}\:{A}\overset{\wedge} {{Y}X} \\ $$$${are}\:{both}\:{less}\:{than}\:\mathrm{90},{they}\:{must}\:{be}\:{equal}. \\ $$$$\Rightarrow\mathrm{60}°+{A}\overset{\wedge} {{Y}X}+{Z}\overset{\wedge} {{Y}B}=\mathrm{90}^{\mathrm{0}} \Rightarrow{A}\overset{\wedge} {{Y}X}={Z}\overset{\wedge} {{Y}B}=\mathrm{15}°, \\ $$$${So},{we}\:{get}\:{Y}\overset{\wedge} {{Z}B}=\mathrm{75}°.{So},\:{sine}\:{rule}\:{in}\:\bigtriangleup{YZB}\:{gives} \\ $$$${the}\:{result}. \\ $$