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How is the angle $\theta$ of the plane and between the $mg$ (attraction due to gravity) and component of $mg$ is same?

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In the figure $\theta$ between the $mg$ and its component $$mg\cos\theta$$ is same as the angle between the horizontal surface and the inclined plane. How?

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closed as off-topic by stafusa, John Rennie, Jon Custer, tpg2114 Aug 12 at 17:33

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  • 1
    $\begingroup$ "Angle between two lines is same as angle between their normals" $\endgroup$ – HS Singh Aug 11 at 5:52
  • $\begingroup$ Thanks, i didn't know that. $\endgroup$ – Dujana Abrar Aug 11 at 5:59
  • $\begingroup$ This should be something proved in a geometry class $\endgroup$ – Kyle Kanos Aug 11 at 11:29
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The triangle between the horizontal, the slope, and the downward gravitational component is a right angle composed of the angles $\theta,90,$ and $90-\theta$, with the $90-\theta$ angle occurring in the top right corner. Because that angle forms a 90 degree angle with the angle in question, then the unknown angle ($\phi$) must be $90-90+\theta = \theta$.

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The triangle made by the incline, ground and mg AND the triangle made by mg , mg cos $\theta$, and the dash line connecting these two vectors are similar.

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