Question Number 148543 by liberty last updated on 29/Jul/21

Answered by Rasheed.Sindhi last updated on 30/Jul/21

Let ∠B=α , ∠C=β , △ABC: AC^2 =AB^2 +BC^2 −2.AB.BC.cos α AC^2 =8^2 +10^2 −2.8.10.cos α =164−160cos α △BCD: BD^2 =10^2 +12^2 −2.10.12cos β =244−240cos β AC^2 +CD^2 =AD^2 =(2R)^2 AB^2 +BD^2 =AD^2 =(2R)^2 Continue

Answered by mnjuly1970 last updated on 29/Jul/21

AC^( 2) =164−160cos(B) AC^( 2) =4R^( 2) −144 164−160cos(B)=4R^( 2) −144 308−4R^( 2) =160cos(B) 308−4R^( 2) =160(−cos(D)) 4R^( 2) −308=160.((12)/(2R))=((960)/R) R^( 3) −77R−240=0

Answered by mr W last updated on 29/Jul/21

sin^(−1) (4/R)+sin^(−1) (5/R)+sin^(−1) (6/R)=(π/2) cos (sin^(−1) (4/R)+sin^(−1) (5/R))=cos ((π/2)−sin^(−1) (6/R)) (√((1−((16)/R^2 ))(1−((25)/R^2 ))))−(4/R)×(5/R)=(6/R) (1−((16)/R^2 ))(1−((25)/R^2 ))=(((20)/R^2 )+(6/R))^2 1=((77)/R^2 )+((240)/R^3 ) R^3 −77R−240=0 R=2(√((77)/3)) sin (((2π)/3)−(1/3)sin^(−1) ((120×3(√3))/(77(√(77))))) ≈10.045

Commented byliberty last updated on 30/Jul/21

great

Answered by mr W last updated on 29/Jul/21

an other way: AC=(√(4R^2 −12^2 )) BD=(√(4R^2 −8^2 )) ABCD is cyclic, therefore AC×BD=AB×CD+BC×AD ⇒(4R^2 −12^2 )(4R^2 −8^2 )=(8×12+20R)^2 ⇒R^3 −77R−240=0 ......

Commented byTawa11 last updated on 03/Aug/21

Great