Question

помогите с геометрией)))))

Answer 1

$frac{BC}{sinA} = frac{AB}{sinC} \ sinC= frac{ABsinA}{BC} \ sinC= frac{ sqrt{6} sin45}{2} = frac{ sqrt{3} }{2} \ C=60$

$CD= sqrt{CE^2+DE^2-2CDcdot DEcosE} = \ = sqrt{CE^2+DE^2-2CEcdot DE sqrt{1-sin^2E}} \ CD= sqrt{13^2+11^2-2cdot13cdot 11 sqrt{1-( frac{12}{13} )^2}} = sqrt{180} = 6sqrt{5}$