# Question-68309

Question Number 68309 by ajfour last updated on 08/Sep/19
Commented by ajfour last updated on 08/Sep/19
$${Find}\:{AB}\:{and}\:{AC}\:{in}\:{terms}\:{of}\:{a}\: \\$$$${and}\:\theta. \\$$
Commented by mr W last updated on 08/Sep/19
$$\frac{{AB}}{\mathrm{sin}\:\theta}=\frac{{AC}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{a}}{\mathrm{sin}\:\mathrm{3}\theta} \\$$$$\Rightarrow{AB}=\frac{{a}\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{{a}}{\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{AC}=\frac{{a}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{\mathrm{2}{a}\:\mathrm{cos}\:\theta}{\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$
Commented by ajfour last updated on 09/Sep/19
$${Thank}\:{you}\:{Sir}.\:{I}\:{shall}\:{find}\:{a} \\$$$${better}\:{question}. \\$$
Answered by ajfour last updated on 09/Sep/19
$${let}\:\:{AC}={ap}\:\:{and}\:{AB}={q} \\$$$$\frac{{ap}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{q}}{\mathrm{sin}\:\theta}\:\:\Rightarrow\:\:{q}=\frac{{ap}}{\mathrm{2cos}\:\theta} \\$$$$\mathrm{cos}\:\left(\pi−\mathrm{3}\theta\right)=\frac{\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4cos}\:^{\mathrm{2}} \theta}+{p}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{2cos}\:\theta}\right)} \\$$$$\Rightarrow\:\left(\mathrm{3}−\mathrm{4cos}\:^{\mathrm{2}} \theta\right){p}^{\mathrm{2}} ={p}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{4cos}\:^{\mathrm{2}} \theta}−\mathrm{1} \\$$$$\Rightarrow\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4cos}\:^{\mathrm{2}} \theta}+\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{2} \\$$$$\:\:\:\:{If}\:\:\:\:\mathrm{0}<\:\theta\:<\:\frac{\pi}{\mathrm{2}} \\$$$$\Rightarrow\:\frac{\mathrm{1}}{{p}}=\frac{\mathrm{1}}{\mathrm{2cos}\:\theta}−\mathrm{2cos}\:\theta\:=\frac{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}{\mathrm{2cos}\:\theta} \\$$$${as}\:\:{AC}={ap}\:\:{and}\:{AB}=\frac{{ap}}{\mathrm{2cos}\:\theta} \\$$$${so}\:\:\boldsymbol{{AC}}=\:\frac{\mathrm{2}{a}\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}\:\:\:;\:\:\boldsymbol{{AB}}=\frac{{a}}{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}\: \\$$