Respuesta :

teamath333
If m ACD = 30 => m DCB = 60.
In triangle ACD:
[tex]sin 30^{o}=sin \frac{AD}{AC} =\ \textgreater \ \frac{1}{2} = \frac{8}{AC} =\ \textgreater \ AC=16 [/tex]
[tex]CD= 8 \sqrt{3} [/tex]
[tex]BC= 16\sqrt{3} [/tex]
AC^{2}+CB^{2}=AB^{2} => AB=[tex] \sqrt{1024} [/tex]=32.
=> P=16+32+[tex]16 \sqrt{3} [/tex]=48 +[tex]16 \sqrt{3} [/tex].
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milanaalixzandar

If m ACD = 30 => m DCB = 60.

In triangle ACD:

AC^{2}+CB^{2}=AB^{2} => AB==32.

=> P=16+32+=48 +.