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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$\begingroup$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.
$\endgroup$ $\begingroup$Picture describes the two sides of the rectangle, multiply together to get area.
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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 $$w = d \sin{\theta}$$ $$l = d \cos{\theta}$$.
$\endgroup$ $\begingroup$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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