Question

repede va rog.............​repedeee.

Answer 1

Răspuns:

Explicație pas cu pas:

a. ∡(BD',(ABC))=∡D'BD

ΔD'BD tr dr

sin ∡D'BD=$\frac{DD'}{BD'}$

BD=12 (diagonala patrat ABCD)

Aplicam pitagora sa aflam DD'

BD'²=DD'²+BD²

225=DD'²+144

DD'=√81=9

sin ∡D'BD=$\frac{9}{15} =\frac{3}{5}$

b. ∡(BD', (ADD'))=∡BD'A

Aflam D'A in ΔAD'D prin pitagora

D'A²=AD²+DD'²

D'A²=72+81=153

D'A=3√17

ΔD'AB este dr in A

sin ∡BD'A=$\frac{AB}{BD'} =\frac{6\sqrt{2} }{15} =\frac{2\sqrt{2} }{5}$