How do you convert the angle $-408^circ16' 20''$ in decimal degree form?

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Answer 1 · 2017-03-16T09:13:13

$= - 408.272^@$

$-408^@16^'20^('')$ = $-408^@ + (16 *60)^('') +20^('')$

$= -408^@ + 980^('')$

$1^@ = 3600^('')$

Therefore,
$-408^@16^'20^('') = -408^@ + (980/3600)^@$

$= -408^@ + 0.272^@$

$= - 408.272^@$

How do you convert the angle -408^circ16' 20''-408^circ16' 20'' in decimal degree form?