Chapter 5.6, Problem 35E

(a)

To determine

To show: That $\frac{a}{b}=\frac{\sin A}{\sin B}$ is true for any $\Delta ABC$ .

Explanation of Solution

Given information: Critical Thinking: $\frac{a}{b}=\frac{\sin A}{\sin B}$

Calculation:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

This is the law of sines. It shows three equations. But any two can be taken by themselves as equal to each other.

  $\frac{a}{\sin A}=\frac{b}{\sin B}$

Here we have the first two. If we cross multiply we get:

  $a\sin B=b\sin A$

Divide both sides by b .

  $\frac{a\sin B}{b}=\frac{b\sin A}{b}\text{ which becomes }\frac{a\sin B}{b}=\sin A$

Now divide both sides by $\sin B$

  $\frac{a\sin B}{b\sin B}=\frac{\sin A}{\sin B}\text{ which simplifies to }\frac{a}{b}=\frac{\sin A}{\sin B}$

Which we are trying to prove

(b)

To determine

To show: That $\frac{a-c}{c}=\frac{\sin A-\sin C}{\sin C}$ is true for any $\Delta ABC$ .

Explanation of Solution

Given information: Critical Thinking: $\frac{a-c}{c}=\frac{\sin A-\sin C}{\sin C}$

Calculation:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

This is the law of sines. It shows three equations. But any two can be taken by themselves as equal to each other.

  $\frac{a}{\sin A}=\frac{c}{\sin C}$

Here we have the first and third. If we cross multiply we get:

  $a\sin C=c\sin A$

Divide both sides by c .

  $\frac{a\sin C}{c}=\frac{c\sin A}{c}\text{ which becomes }\frac{a\sin C}{c}=\sin A$

Now divide both sides by $\sin C$

  $\frac{a\sin C}{c\sin C}=\frac{\sin A}{\sin C}\text{ which simplifies to }\frac{a}{c}=\frac{\sin A}{\sin C}$

This is basically what we proved in part a except we are using c this time instead of b . but we need to go a step further.

We now subtract 1 from both sides.

  $\frac{a}{c}-1=\frac{\sin A}{\sin C}-1$

But one can be written as any quantity over itself.

On the left side I will replace 1 with $\frac{c}{c}$ and on the right side I will replace 1 with $\frac{\sin C}{\sin C}$

  $\frac{a}{c}-\frac{c}{c}=\frac{\sin A}{\sin C}-\frac{\sin C}{\sin C}$

Now since each side has a common denominator they can be written,

  $\frac{a-c}{c}=\frac{\sin A-\sin C}{\sin C}$

Which we are trying to prove

(c)

To determine

To show: That $\frac{a+c}{a-c}=\frac{\sin A+\sin C}{\sin A-\sin C}$ is true for any $\Delta ABC$ .

Explanation of Solution

Given information: Critical Thinking: $\frac{a+c}{a-c}=\frac{\sin A+\sin C}{\sin A-\sin C}$

Calculation:

Trick:

Set $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\left(k\ne 0\right)$

Then:

  $a=k\sin A,b=k\sin B,c=k\sin C$

Apply the rick, we get:

  $\frac{a+c}{a-c}=\frac{k\sin A+k\sin C}{k\sin A-k\sin C}=\frac{\sin A+\sin C}{\sin A-\sin C}$

Which we are trying to prove

(d)

To determine

To show: That $\frac{b}{a+b}=\frac{\sin B}{\sin A+\sin B}$ is true for any $\Delta ABC$ .

Explanation of Solution

Given information: Critical Thinking: $\frac{b}{a+b}=\frac{\sin B}{\sin A+\sin B}$

Calculation:

Trick:

Set $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\left(k\ne 0\right)$

Then:

  $a=k\sin A,b=k\sin B,c=k\sin C$

Apply the rick, we get:

  $\frac{b}{a+b}=\frac{k\sin B}{k\sin A+k\sin B}=\frac{\sin B}{\sin A+\sin B}$

Which we are trying to prove