Reduce the equation sqrt{3} x + y - 8 = 0 int | vedclass

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Question

(A) The given equation is $\sqrt{3} x + y = 8$. Dividing the equation by $\sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$,we get: $\frac{\sqrt{3}}{2}

Vedclass · Accepted Answer

$p = 4, \omega = 30^{\circ}$

Vedclass · Answer

(A) The given equation is $\sqrt{3} x + y = 8$. Dividing the equation by $\sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$,we get: $\frac{\sqrt{3}}{2} x + \frac{1}{2} y = \frac{8}{2}$ $\frac{\sqrt{3}}{2} x + \frac{1}{2} y = 4$. Comparing this with the normal form $x \cos \omega + y \sin \omega = p$,we have: $\cos \omega = \frac{\sqrt{3}}{2}$ and $\sin \omega = \frac{1}{2}$. Since both $\cos \omega$ and $\sin \omega$ are positive,$\omega$ lies in the first quadrant. Thus,$\omega = 30^{\circ}$ and $p = 4$.

Reduce the equation $\sqrt{3} x + y - 8 = 0$ into normal form. Find the values of $p$ and $\omega$.

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