Consider the following statements.I. In trian | vedclass

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(A) For statement $I$: Given $c=6$ and $\cos C=-\frac{11}{25}$.We know $\sin C = \sqrt{1-\cos^2 C} = \sqrt{1-\frac{121}{625}} = \sqrt{\frac{504}{625}}

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(A) For statement $I$: Given $c=6$ and $\cos C=-\frac{11}{25}$.We know $\sin C = \sqrt{1-\cos^2 C} = \sqrt{1-\frac{121}{625}} = \sqrt{\frac{504}{625}} = \frac{\sqrt{36 	imes 14}}{25} = \frac{6\sqrt{14}}{25}$.Using the sine rule,$\frac{c}{\sin C} = 2R$,so $R = \frac{c}{2\sin C} = \frac{6}{2 	imes \frac{6\sqrt{14}}{25}} = \frac{25}{2\sqrt{14}}$.Thus,statement $I$ is true.For statement $II$: Given $a=3, b=4, c=6$.To check if it is acute-angled,we calculate $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9+16-36}{2(3)(4)} = \frac{-11}{24}$.Since $\cos C  90^{\circ})$.Thus,statement $II$ is false.

Consider the following statements.
$I$. In $\triangle ABC$,if $c=6$ and $\cos C=-\frac{11}{25}$,then $R=\frac{25}{2\sqrt{14}}$.
$II$. In $\triangle ABC$,if $a=3, b=4, c=6$,then $\triangle ABC$ is an acute-angled triangle.
Which of the above statements is/are true?

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Consider the following statements.$I$. In $\triangle ABC$,if $c=6$ and $\cos C=-\frac{11}{25}$,then $R=\frac{25}{2\sqrt{14}}$.$II$. In $\triangle ABC$,if $a=3, b=4, c=6$,then $\triangle ABC$ is an acute-angled triangle.Which of the above statements is/are true?