If the coefficient of friction is -3 - the angle of friction-

The correct option is B 60^o We know, #160;coefficient of friction, #160; μ = tan θ ∴√(3) #160;= tan θ⇒θ = 60^o ...

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Applying NLM along 3 Axis

If the coeffi...

Question

If the coefficient of friction is $\sqrt{3}$ , the angle of friction:

30^{o}

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60^{o}

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45^{o}

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37^{o}

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Solution

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30^{o}

({sin}^{2} 30^{o} + {cos}^{2} 30^{o}) - ({sin}^{2} 60^{o} + {cos}^{2} 60^{o}) + ({sin}^{2} 45^{o} + {cos}^{2} 45^{o})

\frac{1 + {tan}^{2} 30^{o}}{1 - {tan}^{2} 30^{o}} + {csc}^{2} 60^{0} - {cos}^{2} 45^{o} + {sin}^{2} 45^{o} + \frac{1 + {cot}^{2} 60^{o}}{1 - {cot}^{2} 60^{o}}

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

\frac{x {cosec}^{2} 30^{o} {sec}^{2} 45^{o}}{8 {cos}^{2} 45^{o} {sin}^{2} 90^{o}} = {tan}^{2} 60^{o} - {tan}^{2} 45^{o}

x

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