Chapter 12.5, Problem 3MCQ

To determine

To calculate : The value of the six trigonometric function for the given triangle.

Answer to Problem 3MCQ

The six trigonometric function are.

$\begin{array}{l}\sin (\theta )=0.66\\ \cos (\theta )=0.75\\ \tan (\theta )=0.88\end{array}$

$\begin{array}{l}\cot (\theta )=1.13\\ \sec (\theta )=1.33\\ \csc (\theta )=1.51\end{array}$

Explanation of Solution

Given information : A diagram of the right-angled triangle is given. Also, adjacent as $9$ and hypotenuse as $12$ is given.

Formula used : For angle $\theta$ in the given triangle,

$\begin{array}{l}\sin (\theta )=\left(\frac{a}{c}\right)\\ \cos (\theta )=\left(\frac{b}{c}\right)\\ \tan (\theta )=\left(\frac{a}{b}\right)\end{array}$ .

And

$\begin{array}{l}\cot (\theta )=\left(\frac{b}{a}\right)\\ \sec (\theta )=\left(\frac{c}{b}\right)\\ \csc (\theta )=\left(\frac{c}{a}\right)\end{array}$

And, we will use the Pythagorean theorem $c^2=a^2+b^2$ . Where, $a$ is the opposite, $b$ is adjacent and $c$ is the hypotenuse of the triangle for the angle $\theta$

Calculation : Find the opposite $a$ of the triangle.

  $\begin{array}{l}{c}^2=a^2+b^2\\ a^2=c^2-b^2\\ a^2={12}^2-9^2\\ a^2=144-81\\ a^2=63\\ \sqrt{a^2}=\sqrt{63}\\ a=7.9\end{array}$

Find the trigonometric ratio

$\begin{array}{l}\sin (\theta )=\left(\frac{a}{c}\right)=\left(\frac{7.9}{12}\right)=0.66\\ \cos (\theta )=\left(\frac{b}{c}\right)=\left(\frac{9}{12}\right)=0.75\\ \tan (\theta )=\left(\frac{a}{b}\right)=\left(\frac{7.9}{9}\right)=0.88\end{array}$

Now,

$\begin{array}{l}\cot (\theta )=\left(\frac{b}{a}\right)=\left(\frac{9}{7.9}\right)=1.13\\ \sec (\theta )=\left(\frac{c}{b}\right)=\left(\frac{12}{9}\right)=1.33\\ \csc (\theta )=\left(\frac{c}{a}\right)=\left(\frac{12}{7.9}\right)=1.51\end{array}$