Chapter 6, Problem 3CLT

To determine

The value of $\cos \theta$ when $\text{tan}\theta \text{=-}\frac{21}{20}\text{and sin}\theta <0$ .

Answer to Problem 3CLT

The value of $\text{cos}\theta \text{ is }\frac{20}{29}$ .

Explanation of Solution

Given info:

  $\text{tan}\theta \text{=-}\frac{21}{20}\text{and sin}\theta <0$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant IIsine functions are positive.

In quadrant IIItangent functions are positive.

In quadrant IVcosine functions are positive.

  

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \text{sin}\theta \text{=}\frac{y}{r}\\ \text{cos}\theta \text{=}\frac{x}{r}\\ \text{tan}\theta \text{=}\frac{y}{x}\\ \text{csc}\theta \text{=}\frac{r}{y}\\ \text{sec}\theta \text{=}\frac{r}{x}\\ \text{cot}\theta \text{=}\frac{x}{y}\end{array}$

Calculation:

Given that $\text{sin}\theta <0$

Hence angle $\theta$ lies inquadrant III or quadrant IV

And $\text{tan}\theta <0$

Hence angle $\theta$ lies inquadrant II or quadrant IV

The common quadrant is quadrant IV.

So $\theta$ lies inquadrant IV.

Now we have

  $\begin{array}{l}\text{tan}\theta \text{=-}\frac{21}{20}=\frac{y}{x}\\ y=-21\\ x=20\\ r=\sqrt{x^2+y^2}\\ r=\sqrt{{20}^2+{\left(-21\right)}^2}=29\\ \text{cos}\theta \text{=}\frac{x}{r}\text{=}\frac{20}{29}\end{array}$

Conclusion:

Thus,the value of $\text{cos}\theta \text{ is }\frac{20}{29}$ ..