The area of the triangle formed by the line x | vedclass

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(A) The given equation of the line is $x \sin \alpha + y \cos \alpha = \sin 2\alpha$.Dividing both sides by $\sin 2\alpha$ (where $\sin 2\alpha = 2 \s

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$\sin 2\alpha$

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(A) The given equation of the line is $x \sin \alpha + y \cos \alpha = \sin 2\alpha$.Dividing both sides by $\sin 2\alpha$ (where $\sin 2\alpha = 2 \sin \alpha \cos \alpha$):$\frac{x \sin \alpha}{2 \sin \alpha \cos \alpha} + \frac{y \cos \alpha}{2 \sin \alpha \cos \alpha} = 1$$\frac{x}{2 \cos \alpha} + \frac{y}{2 \sin \alpha} = 1$This is in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,where the $x$-intercept $a = 2 \cos \alpha$ and the $y$-intercept $b = 2 \sin \alpha$.The area of the triangle formed by the line and the coordinate axes is given by $\Delta = \frac{1}{2} |a| |b|$.$\Delta = \frac{1}{2} |2 \cos \alpha| |2 \sin \alpha| = 2 |\sin \alpha \cos \alpha| = |\sin 2\alpha|$.Assuming the area is positive,the result is $\sin 2\alpha$.

The area of the triangle formed by the line $x \sin \alpha + y \cos \alpha = \sin 2\alpha$ and the coordinate axes is

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