Question

Given: ΔАВС, m∠ACB = 90°,m∠ACD = 30°,AD = 8 cm. Find: CD, Perimeter ΔABC

Please solve!

Answer 1

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

In triangle ACD:

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

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

Answer 2

If m ACD = 30 => m DCB = 60.
In triangle ACD:

`sin 30^(o)=sin (AD)/(AC) =\ \textgreater \ (1)/(2) = (8)/(AC) =\ \textgreater \ AC=16`

`CD= 8 √(3)`

`BC= 16√(3)`
AC^{2}+CB^{2}=AB^{2} => AB=
`√(1024)`=32.
=> P=16+32+
`16 √(3)`=48 +
`16 √(3)`.