Question

If 3sinθ + 4cosθ = 5, then the value of sinθ =​

Answer 1

GIVEN :–

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• $\bf 3 \sin( \theta) + 4 \cos( \theta) = 5$

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TO FIND :–

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• sinθ = ?

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SOLUTION :–

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$\bf \implies 3 \sin( \theta) + 4 \cos( \theta) = 5$

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• We know that –

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$\bf \implies \sin^{2} ( \theta) + \cos^{2} ( \theta) = 1$

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$\bf \implies \cos^{2} ( \theta) = 1 - \sin^{2} ( \theta)$

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$\bf \implies \cos ( \theta) = \sqrt{ 1 - \sin^{2} ( \theta)}$

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• So that –

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$\bf \implies 3 \sin( \theta) + 4 \sqrt{ 1 - \sin^{2} ( \theta)} = 5$

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$\bf \implies 4 \sqrt{ 1 - \sin^{2} ( \theta)} = 5 -3 \sin( \theta)$

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• Square on both sides –

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$\bf \implies 16 \{1 - \sin^{2} ( \theta) \} = \{ 5 -3 \sin( \theta) \}^{2}$

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$\bf \implies 16 \{1 - \sin^{2} ( \theta) \} = \{ 5 -3 \sin( \theta) \}^{2}$

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$\bf \implies 16 - 16 \sin^{2} ( \theta) =25 + 9 \sin^{2} ( \theta) - 30 \sin( \theta)$

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$\bf \implies 25 - 16 + 9 \sin^{2} ( \theta) +16 \sin^{2}( \theta) - 30 \sin( \theta) = 0$

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$\bf \implies 25\sin^{2}( \theta) - 30 \sin( \theta) + 9 = 0$

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$\bf \implies 25\sin^{2}( \theta) - 15 \sin( \theta) - 15 \sin( \theta)+ 9 = 0$

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$\bf \implies 5\sin( \theta) \{5\sin( \theta) - 3\} - 3\{5\sin( \theta) - 3\} = 0$

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$\bf \implies \{5\sin( \theta) - 3\} \{5\sin( \theta) - 3\} = 0$

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$\bf \implies \{5\sin( \theta) - 3\}^{2} = 0$

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$\bf \implies 5\sin( \theta) - 3 = 0$

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$\bf \implies 5\sin( \theta) = 3$

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$\bf \implies \large{ \boxed{ \bf \sin( \theta) = \dfrac{3}{5}}}$

Answer 2

${\sf{\underline{\underline{\pink{GIVEN :–}}}}}$

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• $\bf 3 \sin( \theta) + 4 \cos( \theta)$

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${\sf{\underline{\underline{\pink{To \ FIND :–}}}}}$

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• sinθ = ?

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${\sf{\underline{\underline{\pink{SOLUTION :–}}}}}$

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$\bf \implies 3 \sin( \theta) + 4 \cos( \theta) = 5$

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${\sf{\underline{\underline{\pink{We \ know \ that-}}}}}$

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$\bf \implies \sin^{2} ( \theta) + \cos^{2} ( \theta) = 1$

$\bf \implies \cos^{2} ( \theta) = 1 - \sin^{2} ( \theta)$

$\bf \implies \cos ( \theta) = \sqrt{ 1 - \sin^{2} ( \theta)}$

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${\sf{\underline{\underline{\pink{ So \ that –}}}}}$

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$\bf \implies 3 \sin( \theta) + 4 \sqrt{ 1 - \sin^{2} ( \theta)} = 5$

$\bf \implies 4 \sqrt{ 1 - \sin^{2} ( \theta)} = 5 -3 \sin( \theta)$

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${\sf{\underline{\underline{\pink{ Square \ on \ both \ sides –}}}}}$

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$\bf \implies 16 \{1 - \sin^{2} ( \theta) \} = \{ 5 -3 \sin( \theta) \}^{2}$

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$\bf \implies 16 \{1 - \sin^{2} ( \theta) \} = \{ 5 -3 \sin( \theta) \}^{2}$

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$\bf \implies 16 - 16 \sin^{2} ( \theta) =25 + 9 \sin^{2} ( \theta) - 30 \sin( \theta)$

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$\bf \implies 25 - 16 + 9 \sin^{2} ( \theta) +16 \sin^{2}( \theta) - 30 \sin( \theta) = 0$

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$\bf \implies 25\sin^{2}( \theta) - 30 \sin( \theta) + 9 = 0$

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$\bf \implies 25\sin^{2}( \theta) - 15 \sin( \theta) - 15 \sin( \theta)+ 9 = 0$

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$\bf \implies 5\sin( \theta) \{5\sin( \theta) - 3\} - 3\{5\sin( \theta) - 3\} = 0$

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$\bf \implies \{5\sin( \theta) - 3\} \{5\sin( \theta) - 3\} = 0$

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$\bf \implies \{5\sin( \theta) - 3\}^{2} = 0$

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$\bf \implies 5\sin( \theta) - 3 = 0$

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$\bf \implies 5\sin( \theta) = 3$

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$\bf \implies \large{ \boxed{ \bf \sin( \theta) = \dfrac{3}{5}}}$