Question

if x is equal to 3 sin theta and Y is equal to 4 cos theta then find the value of under root 16 X square + 9 y square?

Answer 1

$\textsf{Given:}$

$\mathsf{x=3\,sin\theta\;\;and}$

$\mathsf{y=4\,cos\theta}$

$\implies\mathsf{\frac{x}{3}=sin\theta\;\&\;\frac{y}{4}=cos\theta}$

$\textsf{We know that}$

$\boxed{\mathsf{sin^2\theta+cos^2\theta=1}}$

$\mathsf{(\frac{x}{3})^2+(\frac{y}{4})^2=1}$

$\mathsf{\frac{x^2}{9}+\frac{y^2}{16}=1}$

$\textsf{simplifying , we get}$

$\mathsf{16x^2+9y^2=9{\times}16}$

$\mathsf{16x^2+9y^2=144}$

$\mathsf{\sqrt{16x^2+9y^2}=\sqrt{144}}$

$\implies\boxed{\sqrt{\mathsf{16x^2+9y^2=12}}}$

Answer 2

Answer:

16

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