Chapter 4.3, Problem 58E

(a)

To determine

Value of $\theta$ in degrees and radians.

Answer to Problem 58E

The angle in degrees is ${60}^0$ and angle in radians is $\frac{\pi }{3}$

Explanation of Solution

Given info:

  $\cot \theta =\frac{\sqrt{3}}{3}$

Formula used:

  $\begin{array}{l}\text{Angle in radian = Angle in degree }\times \frac{\pi }{180}\\ \text{Angle in degree = Angle in radian}\times \frac{180}{\pi }\end{array}$

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Given that $\cot \theta =\frac{\sqrt{3}}{3}$

Angle, $\theta$ when $\cot \theta =\frac{\sqrt{3}}{3}$ is ${60}^0$

Angle in degrees $={60}^0$

We have

  $\begin{array}{l}\text{Angle in radian = Angle in degree }\times \frac{\pi }{180}\\ \text{Angle in radian = 60 }\times \frac{\pi }{180}\\ \text{Angle in radian = }\frac{\pi }{3}\end{array}$

Conclusion:

Thus,the angle in degrees is ${60}^0$ and angle in radians is $\frac{\pi }{3}$

(b)

To determine

Value of $\theta$ in degrees and radians.

Answer to Problem 58E

The angle in degrees is ${45}^0$ and angle in radians is $\frac{\pi }{4}$

Explanation of Solution

Given info:

  $\sec \theta =\sqrt{2}$

Formula used:

  $\begin{array}{l}\text{Angle in radian = Angle in degree }\times \frac{\pi }{180}\\ \text{Angle in degree = Angle in radian}\times \frac{180}{\pi }\end{array}$

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Given that $\sec \theta =\sqrt{2}$

Angle, $\theta$ when $\sec \theta =\sqrt{2}$ is ${45}^0$

Angle in degrees $={45}^0$

We have

  $\begin{array}{l}\text{Angle in radian = Angle in degree }\times \frac{\pi }{180}\\ \text{Angle in radian = 45 }\times \frac{\pi }{180}\\ \text{Angle in radian = }\frac{\pi }{4}\end{array}$

Conclusion:

Thus,the angle in degrees is ${45}^0$ and angle in radians is $\frac{\pi }{4}$