Chapter 6, Problem 48RE

a.

To determine

To convert: The angle from decimal to degree.

Answer to Problem 48RE

The converted degree is ${32.34}^{\circ }$ .

Explanation of Solution

Given information: ${32}^{\circ }{20}^{\text{'}}{35}^{"}$

Calculation: ${32}^{\circ }{20}^{\text{'}}{35}^{"}$

Now,

  $\begin{array}{c}{35}^{"}={\left(\frac{35}{60}\right)}^{\text{'}},{\text{ since, 1}}^{\text{'}}={60}^{"}\\ ={\left(\frac{7}{72}\right)}^{\text{'}}\end{array}$

Therefore,

  $\begin{array}{l}{32}^{\circ }{20}^{\text{'}}{35}^{"}\\ ={32}^{\circ }+{\left(20+\frac{7}{12}\right)}^{\text{'}}\\ ={32}^{\circ }+{\left(\frac{240+7}{12}\right)}^{\text{'}}\\ ={32}^{\circ }+{\left(\frac{247}{12\times 60}\right)}^{\circ }\\ ={\left(32+\frac{247}{12\times 60}\right)}^{\circ }\\ ={\left(32+0.34\right)}^{\circ }\\ ={32.34}^{\circ }\end{array}$

Hence, The converted degree is ${32.34}^{\circ }$ .

b.

To determine

To convert: The angle from decimal to degree.

Answer to Problem 48RE

The conversion is ${63}^{\circ }{10}^{\text{'}}{48}^{"}$

Explanation of Solution

Given information: ${63.18}^{\circ }$

Calculation: ${63.18}^{\circ }$

Now,

  $\begin{array}{l}={63}^{\circ }+{0.18}^{\circ }\\ ={63}^{\circ }+{\left(0.18\times 60\right)}^{\text{'}},{\text{ since, 1}}^{\circ }={60}^{\text{'}}\\ ={63}^{\circ }+{\left(10.8\right)}^{\text{'}}\\ ={63}^{\circ }+{10}^{\text{'}}+{0.8}^{"}\\ ={63}^{\circ }{10}^{\text{'}}+{\left(0.8\times 60\right)}^{"},{\text{ since, 1}}^{\text{'}}={60}^{"}\\ ={63}^{\circ }{10}^{\text{'}}+{48}^{"}\\ ={63}^{\circ }{10}^{\text{'}}{48}^{"}\end{array}$

Hence, The conversion is ${63}^{\circ }{10}^{\text{'}}{48}^{"}$