Two forces A and B are acting at an angle - Their resultant -R- will make an angle - with the force B- such that sin - is equal to

sin α=A sin θR sin α=A sin θ√(A^2+B^2+2AB cosθ) ...

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Standard VI

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Delhi Under The Mughal Rule

Two forces A ...

Question

Two forces A and B are acting at an angle $θ$ . Their resultant 'R' will make an angle $α$ with the force B, such that sin $α$ is equal to

\frac{A + B s i n θ}{\sqrt{A^{2} + B^{2} - 2 A B c o s θ}}

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\frac{B s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}

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\frac{A s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}

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\frac{B c o s θ}{\sqrt{A^{2} + B^{2} - 2 A B c o s θ}}

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Solution

The correct option is C $\frac{A s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}$

$s i n α = \frac{A s i n θ}{R}$

$s i n α = \frac{A s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}$

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Two forces A and B are acting at an angle θ. Their resultant 'R' will make an angle α with the force B, such that sin α is equal to

The correct option is C Asinθ√A2+B2+2ABcosθ sinα=AsinθR sinα=Asinθ√A2+B2+2ABcosθ

Two forces A and B are acting at an angle $θ$ . Their resultant 'R' will make an angle $α$ with the force B, such that sin $α$ is equal to

The correct option is C $\frac{A s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}$

$s i n α = \frac{A s i n θ}{R}$

$s i n α = \frac{A s i n θ}{\sqrt{A^{2} + B^{2} + 2 A B c o s θ}}$