Trigonometry Questions

Question Number 10540 by FilupS last updated on 17/Feb/17

Commented by FilupS last updated on 17/Feb/17

$$\mathrm{All}\:\mathrm{side}\:\mathrm{lenghts}\:=\:{n} \\$$$$\angle{CAE}=\theta \\$$$$\mathrm{0}\leqslant\theta<\frac{\pi}{\mathrm{3}} \\$$$$\: \\$$$$\mathrm{1}.\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{sections} \\$$$$\mathrm{2}.\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{non}\:\mathrm{overlapping}\:\mathrm{area} \\$$$$\:\:\:\:\:\:\mathrm{located}\:\mathrm{at}\:\mathrm{point}\:{C} \\$$$$\mathrm{3}.\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bottom}−\mathrm{most} \\$$$$\:\:\:\:\:\:\mathrm{non}\:\mathrm{overlapping}\:\mathrm{area}\:\left(\mathrm{near}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\right) \\$$

Commented by ajfour last updated on 17/Feb/17

$$\:{which}\:{app}\:{did}\:{u}\:{draw}\:{the}\:{triangles}\:{with}\left(\:{if}\:\:{on}\:{mobile}\right)? \\$$

Commented by FilupS last updated on 17/Feb/17

$$\mathrm{Geogebra} \\$$

Answered by mrW1 last updated on 18/Feb/17

$$\frac{{CF}}{\mathrm{sin}\:\angle{CAF}}=\frac{{CA}}{\mathrm{sin}\:\angle{CFA}} \\$$$$\Rightarrow\frac{{CF}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}=\frac{{n}}{\mathrm{sin}\:\left(\pi−\frac{\theta}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)}=\frac{{n}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\$$$$\Rightarrow{CF}=\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}{n} \\$$$${similarily}: \\$$$$\Rightarrow{CG}=\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}{n} \\$$$$\\$$$$\frac{{GB}}{\mathrm{sin}\:\angle{BAG}}=\frac{{AB}}{\mathrm{sin}\:\angle{AGB}} \\$$$$\Rightarrow\frac{{GB}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}=\frac{{n}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\$$$$\Rightarrow{GB}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}{n} \\$$$$\\$$$${FG}={CG}−{CF}=\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right]{n} \\$$$$\\$$$${EF}={CF} \\$$$${FH}={FG} \\$$$${HD}={GB} \\$$$$\\$$$${A}_{\Delta{AEF}} ={A}_{\Delta{ACF}} =\frac{\mathrm{1}}{\mathrm{2}}×{CF}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\$$$$\\$$$${A}_{\Delta{AFH}} ={A}_{\Delta{AFG}} =\frac{\mathrm{1}}{\mathrm{2}}×{FG}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\$$$${A}_{\Delta{AHD}} ={A}_{\Delta{AGB}} =\frac{\mathrm{1}}{\mathrm{2}}×{GB}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\$$$$\\$$$$\left(\mathrm{1}\right) \\$$$${area}\:{of}\:{overlapping}\:{zone}\:{A}_{\mathrm{1}} \\$$$${A}_{\mathrm{1}} =\mathrm{2}×{A}_{\Delta{AFG}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\$$$$\\$$$$\left(\mathrm{2}\right) \\$$$${A}_{\mathrm{2}} ={A}_{\Delta{CHF}} ={A}_{\Delta{ACF}} −{A}_{\Delta{AFH}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\$$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}\right] \\$$$$\\$$$$\left(\mathrm{3}\right) \\$$$${A}_{\mathrm{3}} ={A}_{\Delta{AGB}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\$$

Commented by FilupS last updated on 17/Feb/17

$${A}\mathrm{wesome}\:\mathrm{job}! \\$$