CD/AD=3/4
CD=3k
AD=4k
AC=√[(3k)²+(4k)²]=√[(9k²+16k²)=√25k²=5k
teorema inaltimii
AD²=BD×DC
16k²=BD× 3k
BD=16k²/3k=16k/3
BC=16k/3+3k=(16k+9k)/3=25k/3
AB²=(BC²-AC²=(25k/3)²-(5k)²=625k²/9-25k²=(625k²-225k²)/9=400k²/9
AB²=40²=1600
400k²/9=1600
4k²/9=16
k²=16×9/4=36
k=6
AC=5k=5×6=30 cm
BC=25k/3=25×6/3=50 cm
Atriunghi ABC=40×30/2=600 cm²
sin B=AC/BC=30/50=3/5