Question

Proszę o rozwiązanie zadania 6.9

Answer 1

|AB|=4+16
|AB|=20
x=4
y=16
h=√(x*y)   wzór
h=√64
h=8
|AC|²=h²+x²
|AC|²=64+4² =64+16 =80=16*5
|AC|=4√5
|BC|²=h²+y²
|BC|² =64+16² =64+256 =320 =64*5
|BC|=8√5
a)
tgα = y/h
tgα =16/8 =2
b)
sin∡ACD = x/|AC|
sin∡ACD =4/4√5 =√5/5
c)
cos∡BAC = sin∡ACD =√5/5