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Question-188463




Question Number 188463 by emilagazade last updated on 01/Mar/23
Answered by mr W last updated on 01/Mar/23
((AO)/(CO))=((sin 30)/(sin 70))  ((BO)/(AO))=((sin 40)/(sin α))  ((CO)/(BO))=((sin (20−α))/(sin 20))  (i)×(ii)×(iii):  1=((sin 30×sin 40×sin (20−α))/(sin 70×sin α×sin 20))  ((sin 20)/(tan α))−cos 20=((sin 70×sin 20)/(sin 30×sin 40))  tan α=(1/(((sin 70)/(sin 30×sin 40))+(1/(tan 20))))  ⇒α=tan^(−1) ((1/(((sin 70)/(sin 30×sin 40))+(1/(tan 20)))))=10°
AOCO=sin30sin70BOAO=sin40sinαCOBO=sin(20α)sin20(i)×(ii)×(iii):1=sin30×sin40×sin(20α)sin70×sinα×sin20sin20tanαcos20=sin70×sin20sin30×sin40tanα=1sin70sin30×sin40+1tan20α=tan1(1sin70sin30×sin40+1tan20)=10°
Commented by universe last updated on 02/Mar/23
sin70×sin α×sin 20 = sin 30×sin 40×sin (20−α)  sin 70×sin α×sin 20 = (1/2)2sin 20×cos 20×sin(20−α)  sin 70×sin α = sin 70×sin (20−α)  sin α = sin (20−α)  α= 20−α  2α = 20   α = 10°
sin70×sinα×sin20=sin30×sin40×sin(20α)sin70×sinα×sin20=122sin20×cos20×sin(20α)sin70×sinα=sin70×sin(20α)sinα=sin(20α)α=20α2α=20α=10°
Commented by mr W last updated on 02/Mar/23
great!
great!
Commented by emilagazade last updated on 02/Mar/23
tanks a lot
tanksalot
Commented by emilagazade last updated on 02/Mar/23
thank you Sir
thankyouSir
Answered by HeferH last updated on 02/Mar/23
Commented by HeferH last updated on 02/Mar/23
20° = 2α  α = 10°
20°=2αα=10°
Commented by mr W last updated on 02/Mar/23
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Commented by emilagazade last updated on 02/Mar/23
Nice, thank you
Nice,thankyou

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