Chapter 13: Q17. (page 706)

Find the values of six trigonometric functions for angleshown below.

Short Answer

Expert verified

The six trigonometric functions are: $\sin θ = \frac{\sqrt{7}}{4}, \csc θ = \frac{4}{\sqrt{7}}, \cos θ = \frac{3}{4}, \sec θ = \frac{4}{3}, \tan θ = \frac{\sqrt{7}}{3}, \cot θ = \frac{3}{\sqrt{7}}$ .

Step by step solution

Step 1. Write down the given information.

The given figure is shown below.

Step 2. Concept used.

The six trigonometric functions are $\sin θ, \csc θ, \cos θ, \sec θ, \tan θ, \cot θ$ .

Let $(x, y)$ be a point in X-Y plane and (r)be the length of a line segment drawn from the origin to the point.

In, $Δ O A B, O A = r, O B = x, A B = y$

Therefore, six trigonometric functions are;

$\begin{matrix} \sin θ = \frac{A B}{O A} = \frac{y}{r}, \csc θ = \frac{1}{\sin θ} = \frac{O A}{A B} = \frac{r}{y} \\ \cos θ = \frac{O B}{O A} = \frac{x}{r}, \sec θ = \frac{1}{\cos θ} = \frac{O A}{O B} = \frac{r}{x} \\ \tan θ = \frac{A B}{O B} = \frac{y}{x}, \cot θ = \frac{1}{\tan θ} = \frac{O B}{A B} = \frac{x}{y} \end{matrix}$

Step 3. Calculation.

From the given figure, the side opposite to angle $(θ)$ is evaluated as:

$\begin{matrix} Opposite = \sqrt{16^{2} - 12^{2}} \\ = 4 \sqrt{7} \end{matrix}$

Now, the six trigonometric functions are evaluated as:

$\begin{matrix} \sin θ = \frac{4 \sqrt{7}}{16} = \frac{\sqrt{7}}{4} \csc θ = \frac{4}{\sqrt{7}} \\ \cos θ = \frac{12}{16} = \frac{3}{4} \sec θ = \frac{4}{3} \\ \tan θ = \frac{\sqrt{7}}{3} \cot θ = \frac{3}{\sqrt{7}} \end{matrix}$