SectionB.2Right Triangle Trigonometry

A right triangle is a triangle containing a 90° angle.

The side opposite to the right angle is called the hypotenuse.

The other two angles add to 90° and are called complementary angles.

The relationship between the sides and angles of a right triangle are given by the three basic trig relations which may be recalled with the mnemonic SOH-COH-TOA.

\begin{align*} \sin\theta \amp= \frac{\textrm{opposite}}{\textrm{hypotenuse}} \amp \cos\theta \amp= \frac{\textrm{adjacent}}{\textrm{hypotenuse}} \amp \tan\theta \amp= \frac{\textrm{opposite}}{\textrm{adjacent}} \end{align*}

and their inverses,

\begin{align*} \theta \amp = \sin^{-1} \left(\frac{\textrm{opposite}}{\textrm{hypotenuse}}\right) \amp \theta \amp = \cos^{-1} \left(\frac{\textrm{adjacent}}{\textrm{hypotenuse}}\right) \amp \theta \amp = \tan^{-1} \left(\frac{\textrm{opposite}}{\textrm{adjacent}}\right) \end{align*}
Facts.

The following statements regarding the trig functions and triangles are always true, and remembering them will help you avoid errors.

• $$\sin\text{,}$$ $$\cos$$ and $$\tan$$ are functions of an angle and their values are unitless ratios of lengths.

• The inverse trig functions are functions of unitless ratios and their results are angles.

• The sine of an angle equals the cosine of its complement and vice-versa.

• The sine and cosine of any angle is always a unitless number between -1 and 1, inclusive.

• The sine, cosine, and tangent of angles between 0 and 90° are always positive.

• The inverse trig functions of positive numbers will always yield angles between 0 and 90°

• The legs of a right triangle are always shorter than the hypotenuse.

• Only right triangles have a hypotenuse.

Hints.

Here are some useful tips for angle calculations

• Take care that your calculator is set in degrees mode for this course.

• Always work with angles between 0° and 90° and use positive arguments for the inverse trig functions.

• Following this advice will avoid unwanted signs and incorrect directions caused because $$\dfrac{-a}{b} = \dfrac{a}{-b}\text{,}$$ and $$\dfrac{a}{b} = \dfrac{-a}{-b}$$ and the calculator can't distinguish between them.