Chapter 2, Problem 16RP

a.

To determine

The supplement and the complement of ${83}^{\circ }$

Answer to Problem 16RP

The supplement is ${97}^{\circ }$ and the complement is $7^{\circ }$

Explanation of Solution

Given: The angle is given which is ${83}^{\circ }$

Concept used: Let angle be x

Here, we are using $\left({180}^{\circ }-x\right)$ to find supplement and $\left({90}^{\circ }-x\right)$ to find the complement. Calculation: Supplement = ${180}^{\circ }-{83}^{\circ }$

  →${97}^{\circ }$

Complement = ${90}^{\circ }-{83}^{\circ }$

  →$7^{\circ }$

Conclusion: Hence, the supplement is ${97}^{\circ }$ and the complement is $7^{\circ }$

b.

To determine

The supplement and the complement of ${42}^{\circ }15\text{'}38"$

Answer to Problem 16RP

The supplement is ${137}^{\circ }44\text{'}22"$ and the complement is ${47}^{\circ }44\text{'}22"$

Explanation of Solution

Given: The angle is given which is ${42}^{\circ }15\text{'}38"$

Concept used: Here, Let angle be x

Here, we are using $\left({180}^{\circ }-x\right)$ to find supplement and $\left({90}^{\circ }-x\right)$ to find the complement.

Calculation: Here, Supplement = $\left({179}^{\circ }59\text{'}60"\right)-\left({42}^{\circ }15\text{'}38"\right)$

  →${137}^{\circ }44\text{'}22"$

Complement = $\left({89}^{\circ }59\text{'}60"\right)-\left({42}^{\circ }15\text{'}38"\right)$

  →${47}^{\circ }44\text{'}22"$

Conclusion: The supplement is ${137}^{\circ }44\text{'}22"$ and the complement is ${47}^{\circ }44\text{'}22"$

c.

To determine

The supplement and the complement of ${97}^{\circ }$

Answer to Problem 16RP

The supplement is ${83}^{\circ }$ and the complement is $-7^{\circ }$

Explanation of Solution

Given: The angle is given which is ${97}^{\circ }$

Concept used: Let angle be x

Here, we are using $\left({180}^{\circ }-x\right)$ to find supplement and $\left({90}^{\circ }-x\right)$ to find the complement. Calculation: Supplement = ${180}^{\circ }-{97}^{\circ }$

  →${83}^{\circ }$

Complement = ${90}^{\circ }-{97}^{\circ }$

  →$-7^{\circ }$

Conclusion: Hence, The supplement is ${83}^{\circ }$ and the complement is $-7^{\circ }$