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

## Mar 21, 2018

$a e = 72.0891$
${\theta}_{a e} = \pi - {\tan}^{-} 1 \left(\frac{72.0881}{0.3893}\right)$

#### Explanation:

$a b = 40$
${\theta}_{a b} = 0$
${x}_{a b} = 40 \cos 0 = 40$
${y}_{a b} = 40 \sin 0 = 0$

$b c = 30$
${\theta}_{b c} = - \frac{\pi}{4}$
${x}_{b c} = 30 \cos \left(- \frac{\pi}{4}\right) = 21.2132$
${y}_{b c} = 40 \sin 0 = - 21.2132$

$c d = 50$
${\theta}_{c d} = \frac{\pi}{3}$
${x}_{c d} = 50 \cos \left(\frac{\pi}{3}\right) = 25$
${y}_{c d} = 50 \sin \left(\frac{\pi}{3}\right) = 43.3013$

$\mathrm{de} = 100$
${\theta}_{\mathrm{de}} = \frac{5 \pi}{6}$
${x}_{\mathrm{de}} = 100 \cos \frac{5 \pi}{6} = - 86.6025$
${y}_{\mathrm{de}} = 100 \sin \frac{5 \pi}{6} = 50$

${x}_{a e} = {x}_{a b} + {x}_{b c} + {x}_{c d} + {x}_{\mathrm{de}}$
$= 40 + 21.2132 + 25 - 86.6025$
${x}_{a e} = - 0.3893$
${y}_{a e} = {y}_{a b} + {y}_{b c} + {y}_{c d} + {y}_{\mathrm{de}}$
$= 0 - 21.2132 + 43.3013 + 50$
${y}_{a e} = 72.0881$

$a e = \sqrt{{x}_{a e}^{2} + {y}_{a e}^{2}}$
$a e = 72.0891$
${\theta}_{a e} = {\tan}^{-} 1 \left({y}_{a e} / \left({x}_{a e}\right)\right)$
$= {\tan}^{-} 1 \left(\frac{72.0881}{-} 0.3893\right)$
${\theta}_{a e} = \pi - {\tan}^{-} 1 \left(\frac{72.0881}{0.3893}\right)$