7.2 Right triangle trigonometry  (Page 2/12)

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.

Evaluating trigonometric functions of angles not in standard position

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

$$\begin{array}{ccc}\hfill \text{sin }\alpha & =\frac{\text{opposite }\alpha }{\text{hypotenuse}}\hfill & =\frac{4}{5}\hfill \\ \hfill \text{cos }\alpha & =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}\hfill & =\frac{3}{5}\hfill \\ \hfill \text{tan }\alpha & =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }\hfill & =\frac{4}{3}\hfill \\ \hfill \text{sec }\alpha & =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }\hfill & =\frac{5}{3}\hfill \\ \hfill \text{csc }\alpha & =\frac{\text{hypotenuse}}{\text{opposite }\alpha }\hfill & =\frac{5}{4}\hfill \\ \hfill \text{cot }\alpha & =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }\hfill & =\frac{3}{4}\hfill \end{array}$$

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Try It

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

$$\begin{array}{ccccccccc}\hfill \text{sin }t& =& \frac{33}{65},\hfill & \hfill \text{cos }t& =& \frac{56}{65},\hfill & \hfill \text{tan }t& =& \frac{33}{56},\hfill \\ \hfill \text{sec }t& =& \frac{65}{56},\hfill & \hfill \text{csc }t& =& \frac{65}{33},\hfill & \hfill \text{cot }t& =& \frac{56}{33}\hfill \end{array}$$

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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 $\mathrm{30°},\mathrm{60°},$ and $\mathrm{45°}.$ Remember, however, that when dealing with right triangles, we are limited to angles between $\mathrm{0°}\text{ and 90°}\text{.}$

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

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