# 7.2 Right triangle trigonometry  (Page 2/12)

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When working with right triangles, keep in mind that the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in [link] . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.

Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. Find the required function:
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

## Evaluating trigonometric functions of angles not in standard position

Using the triangle shown in [link] , evaluate $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha ,\mathrm{cos}\text{\hspace{0.17em}}\alpha ,\mathrm{tan}\text{\hspace{0.17em}}\alpha ,\mathrm{sec}\text{\hspace{0.17em}}\alpha ,\mathrm{csc}\text{\hspace{0.17em}}\alpha ,\text{and}\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\alpha .$

Using the triangle shown in [link] ,evaluate $\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}t,\text{cos}\text{\hspace{0.17em}}t,\text{tan}\text{\hspace{0.17em}}t,\text{sec}\text{\hspace{0.17em}}t,\text{csc}\text{\hspace{0.17em}}t,\text{and}\text{\hspace{0.17em}}\text{cot}\text{\hspace{0.17em}}t.$

## Finding trigonometric functions of special angles using side lengths

It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of $\text{\hspace{0.17em}}30°,60°,$ and $\text{\hspace{0.17em}}45°.\text{\hspace{0.17em}}$ Remember, however, that when dealing with right triangles, we are limited to angles between

Suppose we have a $\text{\hspace{0.17em}}30°,60°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle. The sides have lengths in the relation $\text{\hspace{0.17em}}s,\text{s}\sqrt{3},2s.\text{\hspace{0.17em}}$ The sides of a $\text{\hspace{0.17em}}45°,45°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle, have lengths in the relation $\text{\hspace{0.17em}}s,s,\sqrt{2}s.\text{\hspace{0.17em}}$ These relations are shown in [link] .

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

1. Use the side lengths shown in [link] for the special angle you wish to evaluate.
2. Use the ratio of side lengths appropriate to the function you wish to evaluate.

## Evaluating trigonometric functions of special angles using side lengths

Find the exact value of the trigonometric functions of $\text{\hspace{0.17em}}\frac{\pi }{3},$ using side lengths.

$\begin{array}{ccccc}\hfill \mathrm{sin}\left(\frac{\pi }{3}\right)& =\frac{\text{opp}}{\text{hyp}}\hfill & =\frac{\sqrt{3s}}{2s}\hfill & =\frac{\sqrt{3}}{2}\hfill & \\ \hfill \mathrm{cos}\left(\frac{\pi }{3}\right)& =\frac{\text{adj}}{\text{hyp}}\hfill & =\frac{s}{2s}\hfill & =\frac{1}{2}\hfill & \\ \hfill \mathrm{tan}\left(\frac{\pi }{3}\right)& =\frac{\text{opp}}{\text{adj}}\hfill & =\frac{\sqrt{3}s}{s}\hfill & =\sqrt{3}\hfill & \\ \hfill \mathrm{sec}\left(\frac{\pi }{3}\right)& =\frac{\text{hyp}}{\text{adj}}\hfill & =\frac{2s}{s}\hfill & =2\hfill & \\ \hfill \mathrm{csc}\left(\frac{\pi }{3}\right)& =\frac{\text{hyp}}{\text{opp}}\hfill & =\frac{2s}{\sqrt{3}s}\hfill & =\frac{2}{\sqrt{3}}\hfill & =\frac{2\sqrt{3}}{3}\hfill \\ \hfill \mathrm{cot}\left(\frac{\pi }{3}\right)& =\frac{\text{adj}}{\text{opp}}\hfill & =\frac{s}{\sqrt{3}s}\hfill & =\frac{1}{\sqrt{3}}\hfill & =\frac{\sqrt{3}}{3}\hfill \end{array}$

(x2-2x+8)-4(x2-3x+5)
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