Question

se considera un triunghi dreptunghic ABC in care ipotenuza BC=10, iar sin B=3/5. Sa se determine perimetrul triunghiului.

Answer 1

$BC=10 \\ sinB= \frac{AC}{BC}= \frac{3}{5} \\ \frac{AC}{10}= \frac{3}{5} \\ AC=6 \\ BC^2=AC^2+AB^2 \\ 100=36+AB^2 \\ AB^2=64 \\ AB=8 \\ P=AB+AC+BC=6+8+10=24$

Answer 2

BC=10cm;
sinB=AC/BC=3/5;
⇒AC/10cm=3/5
⇒5AC=30cm⇒AC=6cm;
⇒BC²=AC²+AB²
⇒100cm=36cm+AB²⇒AB²=64⇒AB=8cm;
⇒PΔABC=10cm+6cm+8cm=16cm+8cm=24cm⇒perimetrul dreptunghiului;