John Landry

2022-07-21

Area of rectangle knowing diagonal and angle between diagonal and edge
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 ${d}^{2}\mathrm{cos}\left(\theta \right)\mathrm{sin}\left(\theta \right)$. 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 ?

dominicsheq8

Expert

Step 1
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 $\mathrm{cos}\left(\theta \right)=w/d$ and $\mathrm{sin}\left(\theta \right)=h/d$.
Step 2
If you isolate w and h in these formulas and substitute in the formula "area $=wh$", then the formula you mention appears.

stratsticks57jl

Expert

Step 1
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.
Step 2
Note that we have $\mathrm{sin}\theta =\frac{w}{d}$ and $cos\theta =\frac{l}{d}$. Multiplying by d on both sides for both equations gives $w=d\mathrm{sin}\theta$
$l=d\mathrm{cos}\theta$

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