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Plz solve this?

Asked by dasratna_72 13th May 2017, 7:51 PM
Answered by Expert
Answer:
begin mathsize 16px style tanθ equals fraction numerator 2 straight x left parenthesis straight x plus 1 right parenthesis over denominator 2 straight x plus 1 end fraction equals fraction numerator Opposite space over denominator Adjacent end fraction Hypotenuse equals square root of opposite squared plus height squared end root equals square root of open square brackets 2 straight x left parenthesis straight x plus 1 right parenthesis close square brackets to the power of 2 plus end exponent open parentheses 2 straight x plus 1 close parentheses squared end root rightwards double arrow sinθ equals opposite over hypotenuse equals fraction numerator 2 straight x left parenthesis straight x plus 1 right parenthesis over denominator square root of open square brackets 2 straight x left parenthesis straight x plus 1 right parenthesis close square brackets to the power of 2 plus end exponent open parentheses 2 straight x plus 1 close parentheses squared end root end fraction rightwards double arrow cosθ equals adjacent over hypotenuse equals fraction numerator 2 straight x plus 1 over denominator square root of open square brackets 2 straight x left parenthesis straight x plus 1 right parenthesis close square brackets to the power of 2 plus end exponent open parentheses 2 straight x plus 1 close parentheses squared end root end fraction end style
Answered by Expert 14th May 2017, 9:03 PM
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