A) \[15\times 3\times \sqrt{5}cm\]
B) \[15\times 3\sqrt{7}cm\]
C) \[\frac{15\times \sqrt{7}}{3}cm\]
D) \[\frac{15\times 3}{\sqrt{7}}cm\]
Correct Answer: D
Solution :
Given,\[\mu =\frac{4}{3}\] h = 15 cm R = ? \[\frac{\sin {{90}^{o}}}{\sin C}=\mu \]\[\Rightarrow \]\[\sin C=\frac{1}{\mu }=\frac{R}{\sqrt{{{R}^{2}}+{{h}^{2}}}}=\frac{3}{4}\] \[\Rightarrow \]\[16{{R}^{2}}=9{{R}^{2}}+9{{h}^{2}}\]or,\[7{{R}^{2}}=9{{h}^{2}}\] or\[R=\frac{3}{\sqrt{7}}h=\frac{3}{\sqrt{7}}\times 15cm\]You need to login to perform this action.
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