Math, asked by chandru3011, 1 year ago

answer plzzzzzz ya fast

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Answered by creamiepie
1
Given ,

3cot A = 4

=> cot A = \frac{4}{3}

=> \frac{b}{p} = \frac{4}{3} = k

therefore, b = 4k and p = 3k


➿➿➿➿➿Extra➿➿➿➿➿➿➿
✌✌IF U ARE ASKED TO FIND SIN OR COS VALUE✌✌

APPLYING PYTHAGORAS THEOREM


{h}^{2} = {p}^{2} + {b}^{2}

=> {h}^{2} = {(4k)}^{2} + {(3k)}^{2}

=> {h}^{2} = {16k}^{2} + {9k}^{2}

=> {h}^{2} = {25k}^{2}

=> h = \sqrt{ {25k}^{2}}

=> h = 5k

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Now,

tan A = \frac{3}{4}

again,

 \frac{1 -  {tan}^{2} A}{1 +  {tan}^{2} A}  \\  \\  =   \frac{1 -   { (\frac{3}{4}) }^{2} }{1 + ( { \frac{3}{4} )}^{2} }  \\  \\  =   \frac{1 -  \frac{9}{16} }{1 +  \frac{9}{16 } }  \\  \\  =   \frac{ \frac{16 - 9}{16} }{ \frac{16 + 9}{16} }  \\  \\  =   \frac{ \frac{7}{16} }{ \frac{25}{16} }  \\  \\  =    \frac{7}{16}  \times  \frac{16}{25}  \\  \\  =   \frac{7}{25}



\huge \boxed { \frac{7}{25} }


creamiepie: if it helped u please mark my answer as Brainliest
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