Chapter 9.1, Problem 12E

a.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{9}{15}=\frac{3}{5}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{12}{15}=\frac{4}{5}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{9}{12}=\frac{3}{4}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{12}{9}=\frac{4}{3}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{15}{12}=\frac{5}{4}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{15}{9}=\frac{5}{3}\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(Hypotenous\right)}^2=9^2+{12}^2=225\\ h=15\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{9}{15}=\frac{3}{5}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{12}{15}=\frac{4}{5}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{9}{12}=\frac{3}{4}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{12}{9}=\frac{4}{3}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{15}{12}=\frac{5}{4}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{15}{9}=\frac{5}{3}\end{array}$

b.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{8}{10}=\frac{4}{5}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{6}{10}=\frac{3}{5}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{8}{6}=\frac{4}{3}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{6}{8}=\frac{3}{4}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{10}{6}=\frac{5}{3}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{10}{8}=\frac{5}{4}\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle for triangle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(Hypotenous\right)}^2=6^2+8^2=100\\ h=10\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{8}{10}=\frac{4}{5}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{6}{10}=\frac{3}{5}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{8}{6}=\frac{4}{3}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{6}{8}=\frac{3}{4}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{10}{6}=\frac{5}{3}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{10}{8}=\frac{5}{4}\end{array}$

c.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{5}{7}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{2\sqrt{6}}{7}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{5}{2\sqrt{6}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{2\sqrt{6}}{5}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{7}{2\sqrt{6}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{7}{5}\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle for triangle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(7\right)}^2=5^2+bas{e}^2=\sqrt{49-25}=\sqrt{24}=2\sqrt{6}\\ b=2\sqrt{6}\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{5}{7}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{2\sqrt{6}}{7}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{5}{2\sqrt{6}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{2\sqrt{6}}{5}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{7}{2\sqrt{6}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{7}{5}\end{array}$

d.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{6\sqrt{2}}{9}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{3}{9}=\frac{1}{3}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{6\sqrt{2}}{3}=2\sqrt{2}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{3}{6\sqrt{2}}=\frac{1}{2\sqrt{2}}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{9}{3}=3\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{9}{6\sqrt{2}}=\frac{3}{2\sqrt{2}}\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle for triangle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(9\right)}^2={\left(Perpedicular\right)}^2+3^2=\sqrt{81-9}=\sqrt{72}=6\sqrt{2}\\ p=6\sqrt{2}\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{6\sqrt{2}}{9}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{3}{9}=\frac{1}{3}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{6\sqrt{2}}{3}=2\sqrt{2}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{3}{6\sqrt{2}}=\frac{1}{2\sqrt{2}}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{9}{3}=3\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{9}{6\sqrt{2}}=\frac{3}{2\sqrt{2}}\end{array}$

e.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{10}{18}=\frac{5}{9}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{4\sqrt{14}}{18}=\frac{2\sqrt{14}}{9}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{10}{4\sqrt{14}}=\frac{5}{2\sqrt{14}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{4\sqrt{14}}{10}=\frac{2\sqrt{14}}{5}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{18}{4\sqrt{14}}=\frac{9}{2\sqrt{14}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{18}{10}=\frac{9}{5}\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle for triangle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(18\right)}^2={10}^2+{\left(Base\right)}^2=\sqrt{324-100}=\sqrt{224}=4\sqrt{14}\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{10}{18}=\frac{5}{9}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{4\sqrt{14}}{18}=\frac{2\sqrt{14}}{9}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{10}{4\sqrt{14}}=\frac{5}{2\sqrt{14}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{4\sqrt{14}}{10}=\frac{2\sqrt{14}}{5}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{18}{4\sqrt{14}}=\frac{9}{2\sqrt{14}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{18}{10}=\frac{9}{5}\end{array}$

f.

To determine

Evaluate the six trigonometric functions of the $90-\theta$ angle and describe the relationship.

Answer to Problem 12E

  $\begin{array}{l}\sin \left(90-\theta \right)=\frac{p}{h}=\frac{14}{26}=\frac{7}{13}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{4\sqrt{30}}{26}=\frac{2\sqrt{30}}{13}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{14}{4\sqrt{30}}=\frac{7}{2\sqrt{30}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{4\sqrt{30}}{14}=\frac{2\sqrt{30}}{7}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{26}{4\sqrt{30}}=\frac{13}{2\sqrt{30}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{26}{14}=\frac{13}{7}\end{array}$

The relation is,

  $\begin{array}{l}\sin \left(90-\theta \right)=\cos \theta \\ \tan \left(90-\theta \right)=\cot \theta \\ \sec \left(90-\theta \right)=\csc \theta \end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider,

  

Now evaluate the six trigonometric functions of the $90-\theta$ angle for triangle,

  $\begin{array}{l}{\left(Hypotenous\right)}^2={\left(Perpedicular\right)}^2+{\left(Base\right)}^2\\ {\left(26\right)}^2={14}^2+{\left(Base\right)}^2=\sqrt{676-196}=\sqrt{480}=4\sqrt{30}\\ \sin \left(90-\theta \right)=\frac{p}{h}=\frac{14}{26}=\frac{7}{13}\\ \cos \left(90-\theta \right)=\frac{b}{h}=\frac{4\sqrt{30}}{26}=\frac{2\sqrt{30}}{13}\\ \tan \left(90-\theta \right)=\frac{p}{b}=\frac{14}{4\sqrt{30}}=\frac{7}{2\sqrt{30}}\\ \cot \left(90-\theta \right)=\frac{b}{p}=\frac{4\sqrt{30}}{14}=\frac{2\sqrt{30}}{7}\\ \sec \left(90-\theta \right)=\frac{h}{b}=\frac{26}{4\sqrt{30}}=\frac{13}{2\sqrt{30}}\\ \csc \left(90-\theta \right)=\frac{h}{p}=\frac{26}{14}=\frac{13}{7}\end{array}$

Hence, the relationship has been noticed between first two triangles which is,

  $\begin{array}{l}\sin \left(90-\theta \right)=\cos \theta \\ \tan \left(90-\theta \right)=\cot \theta \\ \sec \left(90-\theta \right)=\csc \theta \end{array}$