a) Ta có \(\widehat{IBK}=90^o-65^o=25^o\)
\(IB=3,8.tan65^o=8,15cm\)
\(BK=\dfrac{3,8}{cos65^o}=9cm\)
b) \(IA=3,8.tan\left(65^o-15^o\right)=4,53cm=>AB=3,62cm\)
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