How is trigonometry used in physics

Trigonometry - application tasks

Sine, cosine and tangent

In right triangles, the sides have the following names:

The hypotenuse is always opposite the right angle. From the unit circle can three trigonometric formulas be derived.

$ \ begin {array} {lllcllr} \ sin \ alpha = \ frac {\ text {opposite side}} {\ text {hypotenuse}} &&& \ cos \ alpha = \ frac {\ text {adjacent}} {\ text {hypotenuse }} &&& \ tan \ alpha = \ frac {\ text {opposite side}} {\ text {adjacent}} \ end {array} $

These are used in factual contexts to determine the side lengths in right triangles.

Application tasks

example 1: Statue of Liberty

The height $ h $ of the Statue of Liberty should be determined if the viewing angle $ \ alpha = 10.5 ^ \ circ $ and the distance to the bank $ a = 500 \ \ text {m} $ are known.

So this is about one Altitude measurement from a distance. Since the hypotenuse is unknown and not relevant, the tangent is used for the calculation:

$ \ begin {array} {llll} \ tan (10.5 ^ \ circ) & = & \ frac {h} {500 \ \ text {m}} & \ vert \ cdot 500 \ \ text {m} \ 92.67 \ \ text {m} & = & h & \ end {array} $

Example 2: Tower

Here, too, a height should be determined. The Tower height but can only be calculated if you first determine length $ a $ using the tangent formula:

$ \ begin {array} {llll} \ tan (32 ^ \ circ) & = & \ frac {a} {50 \ \ text {m}} & \ vert \ cdot 50 \ \ text {m} \ 31, 24 \ \ text {m} & = & a & \ \ end {array} $

Finally, the tower height you are looking for is calculated as follows:

$ h = 31.24 \ \ text {m} + 1.80 \ \ text {m} = 33.04 \ \ text {m} $

Example 3: Height in the equilateral triangle

The triangular height $ g $ can be determined with all three formulas, since all angles ($ \ alpha = 60 ^ \ circ $) as well as all sides ($ a $) are known.

The aim here is to show that all three formulas can be used for calculations and that they deliver the same result for $ g $:

$ \ begin {array} {llll} \ sin (60 ^ \ circ) & = & \ frac {g} {a} & \ 0.866 & = & \ frac {g} {a} & \ vert \ cdot a \ \ 0.866 \ cdot a & = & g & \ end {array} $

$ \ begin {array} {llll} \ cos (30 ^ \ circ) & = & \ frac {g} {a} & \ 0.866 & = & \ frac {g} {a} & \ vert \ cdot a \ \ 0.866 \ cdot a & = & g & \ end {array} $

$ \ begin {array} {llll} \ tan (60 ^ \ circ) & = & \ frac {g} {\ frac {a} {2}} & \ 1.73 & = & g \ cdot \ frac { 2} {a} & \ vert \ cdot a \ 1.73 \ cdot a & = & 2g & \ vert: 2 \ 0.866 \ cdot a & = & g & \ end {array} $