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I found on the web that the area of a rectangle with the diagonal of length $d$, and inner angle (between the diagonal and edge) $\theta$ is $d^2\cos(\theta)\sin(\theta)$. However, I wasn't able to deduce it myself. I tried applying law of sines or generalised Pythagorean theorem but I couldn't derive the area using only the length of the diagonal and the angle between diagonal and edge. How might I get to this result ?

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    $\begingroup$ The side lying against its inner angle has length $d \cos(\theta)$ and the side across the angle has length $d \sin(\theta)$. Multiplying those gives you the result. $\endgroup$ – Tim Huijgens Jul 23 '17 at 18:27
  • $\begingroup$ Just find the length, find the width, then multiply. $\endgroup$ – Ahmed S. Attaalla Jul 23 '17 at 18:27
  • $\begingroup$ The diagonal divides the rectangle in two triangles with an angle of $\pi/2$ each. Apply the definition of $\sin$ to get that $d\sin(\theta)$ is one of the cathetus, and $d\cos(\theta)$ is the other. The catheti are the sides of the rectangle, which area is defined as the product of the lengths of its sides. $\endgroup$ – Ginna Jul 23 '17 at 18:28
  • $\begingroup$ Also it's quite simple to check the result, for example we might have $\sin (\theta)=\frac{L}{d}$ and $\cos(\theta)=\frac{W}{d}$ (or the other way around), in either case we have $d^2\cos (\theta) \sin (\theta)=LW$. $\endgroup$ – Ahmed S. Attaalla Jul 23 '17 at 18:30
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If you use the formulas for sine and cosine in right-angled triangles, the formula can be proved rather easily: If the width and the height of the rectangle are resp. $w$ and $h$, then the formulas say $\cos(\theta)=w/d$ and $\sin(\theta)=h/d$. If you isolate $w$ and $h$ in these formulas and substitute in the formula "area $=wh$", then the formula you mention appears.

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Picture describes the two sides of the rectangle, multiply together to get area. Made with https://start.sketchometry.org/# enter image description here

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Let $ABDC$ be a rectangle, with long sides $AB$ and $CD$ of length $l$ and short sides $AC$ and $BD$ of length $w$. And let $AD$ be a diagonal with length $d$ which makes an angle $\theta$ between $CD$ and $AD$.

Note that we have $\sin{\theta} = \frac{w}{d}$ and $cos{\theta} = \frac{l}{d}$. Multiplying by $d$ on both sides for both equations gives $$w = d \sin{\theta}$$ $$l = d \cos{\theta}$$.

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let $a,b$ the side length of the rectangle, then we have $$A=ab$$ where $$\sin(\theta)=\frac{a}{d}$$ and $$\cos(\theta)=\frac{b}{d}$$

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  • $\begingroup$ Probably a typo but $\frac{b}{d}$ correct? $\endgroup$ – WaveX Jul 23 '17 at 18:37
  • $\begingroup$ yes it was a typo, just corrected, thx $\endgroup$ – Dr. Sonnhard Graubner Jul 23 '17 at 18:40

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