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Question-208913




Question Number 208913 by efronzo1 last updated on 27/Jun/24
Answered by A5T last updated on 27/Jun/24
((sinβ)/6)=(1/(AC));((sin(60−β))/3)=(1/(AC))  ⇒((sin(60−β))/(sinβ))=(1/2)⇒β≈40.8934  ⇒sinβ≈0.654654; we can then find AC=(6/(sinβ))  AC≈9.165151
sinβ6=1AC;sin(60β)3=1ACsin(60β)sinβ=12β40.8934sinβ0.654654;wecanthenfindAC=6sinβAC9.165151
Answered by mr W last updated on 27/Jun/24
∠BCD=90°−α+90°−β=120°  BD^2 =6^2 +3^2 −2×3×6 cos 120°=63  AC=2R=((BD)/(sin ∠BCD))=((√(63))/(sin 120°))                    =2(√(21)) ≈9.165
BCD=90°α+90°β=120°BD2=62+322×3×6cos120°=63AC=2R=BDsinBCD=63sin120°=2219.165
Answered by mr W last updated on 27/Jun/24
Commented by mr W last updated on 27/Jun/24
CE=2×6=12  DE=12+3=15  AD=((15)/( (√3)))=5(√3)  AC=(√((5(√3))^2 +3^2 ))=2(√(21))≈9.165
CE=2×6=12DE=12+3=15AD=153=53AC=(53)2+32=2219.165
Commented by Tawa11 last updated on 27/Jun/24
weldone sirs
weldonesirs

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