Chapter 1.2, Problem 5PSA

a.

To determine

To Calculate: The value of given angle in terms of degrees, minutes and seconds.

Answer to Problem 5PSA

  $\theta =87°10^{″}$

Explanation of Solution

Given:

  $\theta =49°32^{\prime }55^{″}+37°27^{\prime }15^{″}$

Concept Used:

  $\begin{array}{l}1°=60\text{'}\\ 1^{\prime }=60^{″}\end{array}$

Calculation:

As per the given problem

  $\begin{array}{l}\theta =49°32^{\prime }55^{″}+37°27^{\prime }15^{″}\\ =\left(49°+32^{\prime }+55^{″}\right)+\left(37°+27^{\prime }+15^{″}\right)\\ =\left(49°+37°\right)+\left(32^{\prime }+27^{\prime }\right)+\left(55^{″}+15^{″}\right)\\ =86°+59^{\prime }+70^{″}\\ =86°+59^{\prime }+60^{″}+10^{″}\\ =86°+59^{\prime }+1^{\prime }+10^{″}\\ =86°+60^{\prime }+10^{″}\\ =86°+1°+10^{″}\\ =87°+10^{″}\\ \Rightarrow \theta =87°10^{″}\end{array}$

Conclusion:

  $\theta =87°10^{″}$

b.

To determine

To Calculate: The value of given angle in terms of degrees, minutes and seconds.

Answer to Problem 5PSA

  $\theta =82°49^{\prime }$

Explanation of Solution

Given:

  $\theta =123°15^{\prime }-40°26^{\prime }$

Concept Used:

  $\begin{array}{l}1°=60\text{'}\\ 1^{\prime }=60^{″}\end{array}$

Calculation:

As per the given problem

  $\begin{array}{l}\theta =123°15^{\prime }-40°26^{\prime }\\ =\left(123°+15^{\prime }\right)-\left(40°+26^{\prime }\right)\\ =\left(123°-40°\right)+\left(15^{\prime }-26^{\prime }\right)\\ =\left(83°\right)+\left(15^{\prime }-26^{\prime }\right)\\ =\left(82°+1°\right)+\left(15^{\prime }-26^{\prime }\right)\\ =82°+\left(1°+15^{\prime }-26^{\prime }\right)\\ =82°+\left(60^{\prime }+15^{\prime }-26^{\prime }\right)\\ =82°+\left(75^{\prime }-26^{\prime }\right)\\ =82°+49^{\prime }\\ =82°49^{\prime }\\ \Rightarrow \theta =82°49^{\prime }\end{array}$

Conclusion:

  $\theta =82°49^{\prime }$