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4 years ago

6

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

1 answer:

sukhopar [10]4 years ago

8 0

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

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$P(A\cap M)=\frac{5}{8}-\frac{1}{4}==\frac{3}{8}$

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