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|...|-znak pierwiastka
h-wysokość drzewa
a-dł. cienia
Al-kąt alfa 30 st.
tgAl=h/a
tg 30=|3|/3
tg30=h/20,5
|3|/3=h/20,5
h=20,5*|3| /3
w zaokragleniu h=11,84
zad 2
sinAl=|5|/3
sin^2Al +cos^2Al =1
cos^2Al=1-sin^2Al=1-(|5|/3)^2=1- 5/9=4/9
cosAl=2/3
tgAl=sinAl/cosAl=|5|/3 :2/3=|5|/3 * 3/2=|5|/2
ctg-odwrotnosc tg.
ctgAl=2*|5|/5
tg30°=√3/3
x/20,5=√3/3
3x=20,5√3 /:3
x=6,83√3
x=11,84 m
2.
3²=(√5)²+x²
9=5+x²
x²=4
x₁=2
x₂=-2 -->odpada
cosα=2/3, tgα=√5/2, ctgα=2/√5=2√5/5