Surveying: in the surveying map shown below, calculate the distance and direction from point E to point A and estimate the enclosed area?

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Surveying: in the surveying map shown below, calculate the distance and direction from point E to point A and estimate the enclosed area?

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Answer 1 · 2018-03-21T15:41:29

$a e = 72.0891$ $ae=72.0891$
$θ_{a e} = π - {tan}^{- 1} (\frac{72.0881}{0.3893})$ $theta_(ae)=pi-tan^-1(72.0881/0.3893)$

Explanation:

$a b = 40$ $ab=40$
$θ_{a b} = 0$ $theta_(ab)=0$
$x_{a b} = 40 cos 0 = 40$ $x_(ab)=40cos0=40$
$y_{a b} = 40 sin 0 = 0$ $y_(ab)=40sin0=0$

$b c = 30$ $bc=30$
$θ_{b c} = - \frac{π}{4}$ $theta_(bc)=-pi/4$
$x_{b c} = 30 cos (- \frac{π}{4}) = 21.2132$ $x_(bc)=30cos(-pi/4)=21.2132$
$y_{b c} = 40 sin 0 = - 21.2132$ $y_(bc)=40sin0=-21.2132$

$c d = 50$ $cd=50$
$θ_{c d} = \frac{π}{3}$ $theta_(cd)=pi/3$
$x_{c d} = 50 cos (\frac{π}{3}) = 25$ $x_(cd)=50cos(pi/3)=25$
$y_{c d} = 50 sin (\frac{π}{3}) = 43.3013$ $y_(cd)=50sin(pi/3)=43.3013$

$d e = 100$ $de=100$
$θ_{d e} = \frac{5 π}{6}$ $theta_(de)=(5pi)/6$
$x_{d e} = 100 \frac{cos (5 π)}{6} = - 86.6025$ $x_(de)=100cos(5pi)/6=-86.6025$
$y_{d e} = 100 \frac{sin (5 π)}{6} = 50$ $y_(de)=100sin(5pi)/6=50$

$x_{a e} = x_{a b} + x_{b c} + x_{c d} + x_{d e}$ $x_(ae)=x_(ab)+x_(bc)+x_(cd)+x_(de)$
$= 40 + 21.2132 + 25 - 86.6025$ $=40+21.2132+25-86.6025$
$x_{a e} = - 0.3893$ $x_(ae)=-0.3893$
$y_{a e} = y_{a b} + y_{b c} + y_{c d} + y_{d e}$ $y_(ae)=y_(ab)+y_(bc)+y_(cd)+y_(de)$
$= 0 - 21.2132 + 43.3013 + 50$ $=0-21.2132+43.3013+50$
$y_{a e} = 72.0881$ $y_(ae)=72.0881$

$a e = \sqrt{x_{a e}^{2} + y_{a e}^{2}}$ $ae=sqrt(x_(ae)^2+y_(ae)^2)$
$a e = 72.0891$ $ae=72.0891$
$θ_{a e} = {tan}^{- 1} (\frac{y_{a e}}{x_{a e}})$ $theta_(ae)=tan^-1(y_(ae)/(x_(ae)))$
$= {tan}^{- 1} (\frac{72.0881}{- 0.3893})$ $=tan^-1(72.0881/-0.3893)$
$θ_{a e} = π - {tan}^{- 1} (\frac{72.0881}{0.3893})$ $theta_(ae)=pi-tan^-1(72.0881/0.3893)$

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Surveying: in the surveying map shown below, calculate the distance and direction from point E to point A and estimate the enclosed area?

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