Question

Sa se afle lungimea diag AC a unui romb ABCD in carese stie AD=15cm si sinÂ=1/5

Answer 1

AD=15
sinA=$\frac{1}{5}$

sinA1=$\frac{BO}{AB}$
sinA2=$\frac{OD}{AD}$

[tex] \frac{BO}{AB} + \frac{OD}{AD}= \frac{BD}{15} [/tex]
AB=AD

$\frac{BD}{15} = \frac{1}{5}$
BD=3

ABD e triunghi isoscel, deci AO il imparte pe BD in jumatate

aplicam teorema lui Pitagora in triunghiul dreptunghic AOB

$1,5^{2}+ AO^{2}= 15^{2}$
AO=10⇒OC=10 

AC=AO+OC
AC=10·2
AC=20