Add the following:a) $ 16^{\circ} 45^{\prime} 35^{\prime \prime} $ and $ 112^{\circ} 42^{\prime} 48^{\prime \prime} $
b) $ 78^{\circ} 19^{\prime} 56^{\prime \prime} ; 95^{\circ} 48^{\prime} 17^{\prime \prime} $ and $ 114^{\circ} $ 34' 41"

Given:

a) \( 16^{\circ} 45^{\prime} 35^{\prime \prime} \) and \( 112^{\circ} 42^{\prime} 48^{\prime \prime} \)
b) \( 78^{\circ} 19^{\prime} 56^{\prime \prime} ; 95^{\circ} 48^{\prime} 17^{\prime \prime} \) and \( 114^{\circ} \) 34' 41"

To do:

We have to add the given terms.

Solution:

We know that,

$1^{\circ}=60^{\prime}$

$1^{\prime}=60^{\prime \prime}$

 Therefore,

a) $16^{\circ} 45^{\prime} 35^{\prime \prime} + 112^{\circ} 42^{\prime} 48^{\prime \prime}=(16+112)^{\circ} (45+42)^{\prime} (35+48)^{\prime \prime}$

$=128^{\circ} 87^{\prime} 83^{\prime \prime}$

$=128^{\circ} (60+27)^{\prime} (60+23)^{\prime \prime}$

$=129^{\circ} (27+1)^{\prime} 23^{\prime \prime}$

$=129^{\circ} 28^{\prime} 23^{\prime \prime}$

b) $78^{\circ} 19^{\prime} 56^{\prime \prime} + 95^{\circ} 48^{\prime} 17^{\prime \prime} + 114^{\circ} 34^{\prime} 41^{\prime \prime}$

$=(78+95+114)^{\circ} (19+48+34)^{\prime} (56+17+41)^{\prime \prime}$

$=287^{\circ} 101^{\prime} 114^{\prime \prime}$

$=287^{\circ} (60+41)^{\prime} (60+54)^{\prime \prime}$

$=(287+1)^{\circ} (41+1)^{\prime} 54^{\prime \prime}$

$=288^{\circ} 42^{\prime} 54^{\prime \prime}$

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Simply Easy Learning

Updated on: 10-Oct-2022

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