Question-23226

Question Number 23226 by ajfour last updated on 27/Oct/17

Commented by ajfour last updated on 27/Oct/17

$${Q}.\mathrm{23212}\:\:\left({solution}\right) \\ $$

Commented by math solver last updated on 28/Oct/17

sir how you write the angles 75 , 60. ahh, i forgot ′:(

$$\mathrm{sir}\:\mathrm{how}\:\mathrm{you}\:\mathrm{write}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{75}\:,\:\mathrm{60}. \\ $$$$\mathrm{ahh},\:\mathrm{i}\:\mathrm{forgot}\:':\left(\right. \\ $$

Commented by ajfour last updated on 28/Oct/17

AB is ⊥ to x axis ∠DAF=60° , ∠FAx =∠DAx−∠DAF =90°−60° =30° ∠BAF=2∠DAF=120° ∠BAx=∠BAF+∠FAx =120°+30°=150° ∠RAx=∠BAx−∠BAR =150°−90°=60° ∠PAF=45° (angle b/w diagonal of square and its side) ∠PAx=∠PAF+∠FAx = 45°+30° = 75° .

$${AB}\:{is}\:\bot\:{to}\:{x}\:{axis} \\ $$$$\angle{DAF}=\mathrm{60}°\:\:,\:\: \\ $$$$\angle{FAx}\:=\angle{DAx}−\angle{DAF} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{90}°−\mathrm{60}°\:=\mathrm{30}° \\ $$$$\angle{BAF}=\mathrm{2}\angle{DAF}=\mathrm{120}° \\ $$$$\angle{BAx}=\angle{BAF}+\angle{FAx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{120}°+\mathrm{30}°=\mathrm{150}° \\ $$$$\angle{RAx}=\angle{BAx}−\angle{BAR} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{150}°−\mathrm{90}°=\mathrm{60}° \\ $$$$\angle{PAF}=\mathrm{45}°\:\:\:\left({angle}\:{b}/{w}\:\:{diagonal}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{square}\:{and}\:{its}\:{side}\right) \\ $$$$\angle{PAx}=\angle{PAF}+\angle{FAx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{45}°+\mathrm{30}°\:=\:\mathrm{75}°\:\:. \\ $$

Commented by math solver last updated on 28/Oct/17

$$\left.\mathrm{alright}\:,\:\mathrm{thanks}\::\right) \\ $$

Answered by ajfour last updated on 27/Oct/17

Let A be the origin. AB=1 so AP =(√2) (diagonal of sq) AR=AB=1 x_P =(√2)cos 75° ; y_P =(√2)sin 75° x_R =cos 60° ; y_R =sin 60° △_(APQ) =(1/2)×2x_P ×y_P =2cos 75°×sin 75° =sin 150° = sin 30° =(1/2) △_(SRP) =(1/2)×2x_R ×(y_P −y_R ) =cos 60°×((√2)sin 75°−sin 60°) =(1/2)[(√2)×((((√3)+1))/(2(√2)))−((√3)/2)] =(1/4) So, (△_(APQ) /△_(SRP) ) =(((1/2))/((1/4))) =2 .

$${Let}\:{A}\:{be}\:{the}\:{origin}. \\ $$$${AB}=\mathrm{1}\:\:{so}\:{AP}\:=\sqrt{\mathrm{2}}\:\:\left({diagonal}\:{of}\:{sq}\right) \\ $$$${AR}={AB}=\mathrm{1} \\ $$$${x}_{{P}} =\sqrt{\mathrm{2}}\mathrm{cos}\:\mathrm{75}°\:\:;\:\:{y}_{{P}} =\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{75}° \\ $$$${x}_{{R}} =\mathrm{cos}\:\mathrm{60}°\:;\:\:{y}_{{R}} =\mathrm{sin}\:\mathrm{60}° \\ $$$$\:\bigtriangleup_{{APQ}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{x}_{{P}} ×{y}_{{P}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\:\mathrm{75}°×\mathrm{sin}\:\mathrm{75}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{sin}\:\mathrm{150}°\:\:=\:\mathrm{sin}\:\mathrm{30}°\:=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\bigtriangleup_{{SRP}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{x}_{{R}} ×\left({y}_{{P}} −{y}_{{R}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cos}\:\mathrm{60}°×\left(\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{75}°−\mathrm{sin}\:\mathrm{60}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{2}}×\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${So},\:\:\:\:\:\:\frac{\bigtriangleup_{{APQ}} }{\bigtriangleup_{{SRP}} }\:=\frac{\left(\mathrm{1}/\mathrm{2}\right)}{\left(\mathrm{1}/\mathrm{4}\right)}\:=\mathrm{2}\:. \\ $$

Commented by math solver last updated on 27/Oct/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$

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