Menu Close

Question-208581




Question Number 208581 by Tawa11 last updated on 18/Jun/24
Commented by mr W last updated on 18/Jun/24
Commented by Tawa11 last updated on 18/Jun/24
Thanks sir.
Thankssir.
Answered by A5T last updated on 18/Jun/24
Commented by A5T last updated on 18/Jun/24
2^2 −2×2×5cosθ=+2×3×4cosθ  ⇒cosθ=(1/(11))  AB=(√(2^2 +5^2 −2×2×5×(1/(11))))=((√(3289))/(11))  ((abc)/(4R))=(1/2)×2×5×((√(120))/(11))⇒R=((√(3289))/( 2(√(120))))
222×2×5cosθ=+2×3×4cosθcosθ=111AB=22+522×2×5×111=328911abc4R=12×2×5×12011R=32892120
Commented by Tawa11 last updated on 18/Jun/24
Thanks sir
Thankssir

Leave a Reply

Your email address will not be published. Required fields are marked *