Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Commented by ajfour last updated on 27/Oct/17

Q.23212  (solution)

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

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

alright , thanks :)

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

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

Thank you sir !

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com