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

Question

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

in progress 0
Raelynn 1 month 2021-08-12T16:51:36+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-12T16:52:46+00:00

    GIVEN :

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

    TO FIND :

    • sinθ = ?

    SOLUTION :

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

    • We know that –

     \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)}

    • So that –

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

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

    • Square on both sides –

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

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

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

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

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

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

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

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

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

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

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

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

    0
    2021-08-12T16:52:52+00:00

    {\sf{\underline{\underline{\pink{GIVEN :-}}}}}

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

    {\sf{\underline{\underline{\pink{To \  FIND :-}}}}}

    • sinθ = ?

    {\sf{\underline{\underline{\pink{SOLUTION :-}}}}}

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

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

    \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)}

    {\sf{\underline{\underline{\pink{ So \ that -}}}}}

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

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

    {\sf{\underline{\underline{\pink{ Square \  on \  both \ sides -}}}}}

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

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

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

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

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

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

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

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

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

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

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

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

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )