Chapter 9, Problem 6REAE

Solution

To find: Evaluate the six trigonometric functions of angle $B$.

Hence, $\sin \angle B=\frac{5}{\sqrt{41}}\cos \angle B=\frac{4}{\sqrt{41}},\tan \angle B=\frac{5}{\sqrt{41}},\sec \angle B=\frac{\sqrt{41}}{5},\csc \angle B=\frac{\sqrt{41}}{5},\cot \angle B=\frac{4}{5}$.

Given: $\text{In}\Delta ABC,a=4,b=5,\text{and}c=90°$

Concept Used: Trigonometric ratio identity

Calculation:

From the figure:

  

  $\text{In}\Delta ABC,a=4,b=5,\text{and}c=90°$

Using Pythagoras theorem:

  $\begin{array}{c}A{B}^2=B{C}^2+A{C}^2\\ A{B}^2=5^2+4^2\\ AB=\sqrt{25+16}\\ AB=\sqrt{41}\end{array}$

Using trigonometric ratio:

  $\begin{array}{l}\sin \angle B=\frac{AC}{AB}\\ \sin \angle B=\frac{5}{\sqrt{41}}\\ \cos \angle B=\frac{BC}{AB}\\ \cos \angle B=\frac{4}{\sqrt{41}}\\ \tan \angle B=\frac{AC}{BC}\\ \tan \angle B=\frac{5}{\sqrt{41}}\\ \sec \angle B=\frac{AB}{BC}\\ \sec \angle B=\frac{\sqrt{41}}{5}\\ \csc \angle B=\frac{AB}{AC}\\ \csc \angle B=\frac{\sqrt{41}}{5}\\ \cot \angle B=\frac{BC}{AC}\\ \cot \angle B=\frac{4}{5}\end{array}$

Conclusion:

Hence, $\sin \angle B=\frac{5}{\sqrt{41}}\cos \angle B=\frac{4}{\sqrt{41}},\tan \angle B=\frac{5}{\sqrt{41}},\sec \angle B=\frac{\sqrt{41}}{5},\csc \angle B=\frac{\sqrt{41}}{5},\cot \angle B=\frac{4}{5}$