Calculate the reduced bearing of a line if its whole circle bearing is 130 degrees.

$50^{0}$

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$S 30^{0} E$

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$S 40^{0} W$

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$N 130^{0} E$

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Solution

The correct option is B
$S 30^{0} E$

Whole circle bearing, $ϕ = 130^{0}$
For quadrant "SE"
Reduced bearing = $S (180^{0} - 130^{0}) E = S 50^{0} E$

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