Question

Answer 1

\(a,EF=\sqrt{DE^2+DF^2}=10\left(cm\right)\\ DK=\dfrac{DE\cdot DF}{EF}=4,8\left(cm\right)\\ b,\sin E=\dfrac{DF}{EF}=\dfrac{4}{5}\approx\sin53^0\\ \Rightarrow\widehat{E}\approx53^0\)

Answer 2

a: EF=10cm

DK=4,8cm

Answer 3