Question

A medida do lado bc do triangulo

Answer 1

Sen60° = h/2
0,87 = h/2
h = 1,74
tg45° = h/DC
1 = 1,74/DC
DC = 1,74
Agora DB
2² = 1,74² + DB²
4 = 3,0276 + DB²
DB² = 4 - 3,0276
DB² = 0,9724
DB = √0,9724
DB = 0,99
BC = DB + DC
BC = 0,99 + 1,74
BC = 2,73m