Using trigonometric relationships

At trigonometric relations are formulas that relate the angles and sides of a right triangle. These formulas involve the functions sine, cosine and tangentand have many applications in geometric problems involving this type of triangle.

Trigonometric relations in the right triangle

O right triangle it is the triangle that has a right angle (90°) and two acute angles (less than 90°). The sides of the right triangle are called the hypotenuse and sides, and the sides can be opposite or adjacent, depending on the reference angle.

rectangle triangle

Elements of the right triangle:

Hypotenuse: side opposite right angle;
Opposite side: side opposite the considered acute angle;
Adjacent side: side consecutive to the considered acute angle.

Formulas:

considering the angle $\dpi{120} \alpha$ of the right triangle, we have to:

$\dpi{120} \mathbf{sen\, \boldsymbol{\alpha} = \frac{catheto\, opposite}{hypotenuse}}$

$\dpi{120} \mathbf{cos\, \boldsymbol{\alpha} = \frac{catheto\, adjacent}{hypotenuse}}$

$\dpi{120} \mathbf{tan\, \boldsymbol{\alpha} = \frac{side\, opposite}{side \, adjacent}}$

Note: The hypotenuse of the right triangle is always the same, the opposite and adjacent sides vary in relation to the acute angle under consideration.

Examples - Using Trigonometric Relationships

Below are examples of how to use trigonometric relationships.

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Example 1: Calculate the value of x and y in the triangle below:

triangle

From the sine of the 30° angle, we can determine the value of x, which is the hypotenuse of the triangle.

$\dpi{120} \mathrm{sen\, 30^{\circ} =\frac{5}{x}}$

$\dpi{120} \Rightarrow \mathrm{ x=\frac{5}{sen\, 30^{\circ}}}$

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$\dpi{120} \mathrm{\Rightarrow x = 10}$

Now, one of the ways to find the value of y is from the cosine of the 30° angle. In this case, y is the leg adjacent to the 30° angle.

$\dpi{120} \mathrm{cos\, 30^{\circ} =\frac{y}{10}}$

$\dpi{120} \Rightarrow \mathrm{ y = 10\cdot cos\, 30^{\circ}}$

$\dpi{120} \Rightarrow \mathrm{ y \approx 9}$

Example 2: Determine the measure of the angles $\dpi{120} \alpha$ and $\dpi{120} \beta$ from the triangle below:

triangle

First, let's determine the angle $\dpi{120} \alpha$ :

$\dpi{120} \mathrm{sen\, \alpha = \frac{5}{6,4}}$

$\dpi{120} \mathrm{\Rightarrow \alpha = sen^{-1} \left ( \frac{5}{6,4}\right )}$

$\dpi{120} \mathrm{\Rightarrow \alpha \approx 51.37^{\circ}}$

Now let's determine the angle $\dpi{120} \beta$ :

$\dpi{120} \mathrm{sen\, \beta = \frac{4}{6,4}}$

$\dpi{120} \mathrm{\Rightarrow \beta = sen^{-1} \left ( \frac{4}{6,4}\right )}$

$\dpi{120} \Rightarrow \beta \approx 38.68$

Note that we used sine in both cases, but we could also use cosine and arrive at these same results.

You may also be interested:

trigonometric table
trigonometric circle
Derived relationships
List of Trigonometry Exercises
Sine and Cosine of Obtuse Angles

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