In Delta ABC,AB = 6sqrt{3} text{ cm},AC = 12 | vedclass

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Question

(B) Given that,$AB = 6\sqrt{3} 	ext{ cm}$,$AC = 12 	ext{ cm}$,and $BC = 6 	ext{ cm}$.Calculating the squares of the sides:$AB^2 = (6\sqrt{3})^2 = 3

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Given that,$AB = 6\sqrt{3} 	ext{ cm}$,$AC = 12 	ext{ cm}$,and $BC = 6 	ext{ cm}$.Calculating the squares of the sides:$AB^2 = (6\sqrt{3})^2 = 36 	imes 3 = 108$$BC^2 = (6)^2 = 36$$AC^2 = (12)^2 = 144$Now,observe that:$AB^2 + BC^2 = 108 + 36 = 144$Since $AB^2 + BC^2 = AC^2$,the triangle satisfies the converse of the Pythagoras theorem.Therefore,$	riangle ABC$ is a right-angled triangle,with the right angle opposite to the hypotenuse $AC$.Thus,$\angle B = 90^\circ$.Hence,the correct answer is $(B)$.

In $\Delta ABC$,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$. The angle $B$ is (in $^o$):

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