Chapter 15.1, Problem 19PSC

a.

To determine

To compare: the complement of $\angle A$ with the complement of $\angle B$ .

Answer to Problem 19PSC

  $\text{complement of}\angle A<\text{complement of}\angle B$ .

Explanation of Solution

Given:

  $\angle A$ is greater than its complement, and the complement of $\angle A$ is greater than complement of $\angle B$ .

Calculation:

According to the question,

  $\begin{array}{l}\angle A>\text{complement of}\angle A\left\{\text{given}\right\}\\ \text{complement of}\angle A>\angle B\left\{\text{given}\right\}\\ \angle A>\angle B\\ \therefore 90-\angle A<90-\angle B\{\text{by special suntraction}}\\ \therefore \text{complement of}\angle A<\text{complement of}\angle B\end{array}$

Hence, $\text{complement of}\angle A<\text{complement of}\angle B$

b.

To determine

To compare: the complement of $\angle B$ with the complement of $\angle A$ .

Answer to Problem 19PSC

  $\text{complement of}\angle B>\angle A$ .

Explanation of Solution

Given:

  $\angle A$ is greater than its complement, and the complement of $\angle A$ is greater than complement of $\angle B$ .

Calculation:

According to the question,

  $\begin{array}{l}\text{complement of}\angle A>\angle B\left\{\text{given}\right\}\\ \text{then,}90-\angle A<\angle B\{\text{by special subtraction}}\\ \Rightarrow 90-\left(90-\angle A\right)<90-\angle B\{\text{by special subtraction}}\\ \angle A<90-\angle B\\ \therefore \angle A<\text{complement of}\angle B,\\ \text{complement of}\angle B>\angle A\end{array}$

Hence, $\text{complement of}\angle B>\angle A$

c.

To determine

To list: $\angle A\text{and}\angle B$ and their complement in order of size from largest to smallest.

Answer to Problem 19PSC

  $\text{compt}\angle B,\angle A,\text{comp}\angle A,\angle B$ .

Explanation of Solution

Given:

  $\angle A$ is greater than its complement, and the complement of $\angle A$ is greater than complement of $\angle B$ .

Calculation:

According to the question,

From part (a), $\text{complement of}\angle A<\text{complement of}\angle B$ .

From part (b), $\text{complement of}\angle B>\angle A$ .

Thus, it is clearly seen that from both parts, the orders of size from largest to smallest are $\text{compt}\angle B,\angle A,\text{comp}\angle A,\angle B$ .

Hence, $\text{complement of}\angle B>\angle A$