Question

$in \: the \: triangle \: abc \: ab = 6 \sqrt{3} \: \: \: and \: ac = 12cm \: and \: bc \: = 6cm \: \: \: find \: the \: angle \: b$

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Answer 1

Answer:

6

Step-by-step explanation:

12^2-6√3^2 = ac^2

144 - 108 =36

ac=√36

ac=6

Answer 2

AB=6√3

AC=12

BC=6

according to properties of triangle

cos B = (AB^2 + BC^2 - AC^2) / 2(AB)(BC)

cosB=((6√3)^2+(6)^2-(12)^2)/(2(6√3)(6)

CosB=(108+36-144)/72√3

cosB=0

B=90°

hope this helps you

please mark the answer as brainlest please