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Mathematics-Online problems:

Solution to the problem of the (previous) week


Problem:

The positions of two points $ X$ and $ Y$ which are difficult to access can be determined by measuring angles.

\includegraphics[width=.7\linewidth]{TdM_12_A3_bild}

Find


Answer:

$ X = $ $ \big($ , $ \big)$  
$ Y = $ $ \big($ , $ \big)$  
$ \vert\overline{XY}\vert = $  
$ F = $  

(The results should be correct to four decimal places.)


Solution:

Since $ \alpha+\beta=90^\circ$, the line segments $ \overline{AY}$ and $ \overline{BX}$ intersect perpendicularly at point $ S$. Then, we can successively determine the lengths of the line segments:
$\displaystyle \vert\overline{AS}\vert$ $\displaystyle =$ $\displaystyle \vert\overline{AB}\vert \cos\alpha = 50\sqrt{3}\,,$  
$\displaystyle \vert\overline{AX}\vert$ $\displaystyle =$ $\displaystyle \dfrac{\vert\overline{AS}\vert}{\cos\delta} = \dfrac{50\sqrt{3}}{\frac12 \sqrt{2+\sqrt{3}}}
= 100\sqrt{6-3\sqrt{3}}\,,$  
$\displaystyle \vert\overline{SX}\vert$ $\displaystyle =$ $\displaystyle \vert\overline{AX}\vert\sin\delta = 50\sqrt{3}\cdot \dfrac12 \sqrt{2-\sqrt{3}}
= 50\sqrt{21-12\sqrt{3}}\,,$  
$\displaystyle \vert\overline{BS}\vert$ $\displaystyle =$ $\displaystyle \vert\overline{AB}\vert\sin\alpha = 100\cdot \dfrac12 = 50\,,$  
$\displaystyle \vert\overline{SY}\vert$ $\displaystyle =$ $\displaystyle \vert\overline{BS}\vert\tan\gamma = 50\sqrt{3}\,,$  
$\displaystyle \vert\overline{XY}\vert$ $\displaystyle =$ $\displaystyle \sqrt{\vert\overline{SX}\vert^2+\vert\overline{SY}\vert^2} = 100\sqrt{6-3\sqrt{3}}\approx 89.6575 \,.$  

So, the area $ F$ of the shaded region is

$\displaystyle F = \dfrac12 \vert\overline{SX}\vert\ \vert\overline{SY}\vert
= ...
...sqrt{9\cdot 7-9\cdot 4\sqrt{3}}
= 3750\sqrt{7-4\sqrt{3}} \approx 1004.8095 \,,
$

and for the coordinates of $ X$ and $ Y$ we obtain

$\displaystyle X$ $\displaystyle =$ $\displaystyle \vert\overline{AX}\vert \big( \cos(\alpha+\delta),\, \sin(\alpha+\delta) \big)$  
  $\displaystyle =$ $\displaystyle 100\sqrt{6-3\sqrt{3}} \left( \dfrac{\sqrt2}{2}, \dfrac{\sqrt2}2 \right)$  
  $\displaystyle =$ $\displaystyle \left( 50\sqrt{12-6\sqrt{3}},\, 50\sqrt{12-6\sqrt{3}} \right)$  
  $\displaystyle \approx$ $\displaystyle (63.3975,\, 63.3975) \,,$  
       
$\displaystyle Y$ $\displaystyle =$ $\displaystyle (\vert\overline{AS}\vert+\vert\overline{SY}\vert) \big( \cos(\alpha),\, \sin(\alpha) \big)$  
  $\displaystyle =$ $\displaystyle 100\sqrt{3} \left( \dfrac{\sqrt3}2,\, \dfrac12 \right)$  
  $\displaystyle =$ $\displaystyle \left( 150,\, 50\sqrt{3} \right)$  
  $\displaystyle \approx$ $\displaystyle (150,\, 86.6025)
\,.$  


[problem of the week]