The tan function

Geometrical methods are very often insufficient for answering questions with a required accuracy. Drawings can only be executed with a finite accuracy and extraction of values from them generates further errors. For this reason, during land surveillance, astronomical predictions, etc., people tried already very early on to obtain results by calculation of more exact values. These methods were improved and led to trigonometry.

As the word trigonometry suggests, as in geometry, figures into which every pattern, bounded by straight lines, can be decomposed - triangles - have a dominant role. If you draw a triangle according to one of the four basic tasks, you learn that 3 items determine an entire triangle and the remaining sides and angles depend on those given. In order to be able to compute the remaining items, you have to formulate algebraically the functional relations between the components of triangles. You have already met a few of these. The rule that the sum of the angles of a triangle equals 180º tells you that every angle is a function of the other two angles, so that, for example, every pair of angles implies a definite angle , computed from the equation Pythagoras' theorem is a relation between the sides of right-angled triangles: ; every side of a right-angled triangle is a function of the other two sides and can be computed, if these are known. In order to clarify the interdependence of the sides and angles of a right-angled triangle, the following approach can be followed.

If you point at a tower a telescope , which can be rotated about a horizontal axis, you can focus it at any angle. As increases, while b₁ stays constant, a₁ increases and vice versa. If you change b₁ - increase the distance between the telescope and the tower - you find out that with the same angle the magnitude of a also depends on b, so that b enters into the relationship between and a. However, you realize that, by the second similarity rule, if the two right-angled triangles A₁B₁C₁ and A₂B₂C₂, are similar, that is, all ratios of their sides remain invariant.

If you select an arbitrary ratio, say a : b, then, if , also a₁ : b₁ = a₂ : b₂ corresponds to a certain angle , that is, a certain ratio a : b belongs to it. As increases, you can see from the figures that also a : b increases and vice versa, that is a/b is a function of , for which you can write This function is called tangent (tan) and is defined for the right-angled triangle by

Of course, it is assumed here that a and b are expressed in the same unit.

Here In order to get tan 31º, since all right-angled triangles with the angle 31º are similar, you can use any of the triangles to find a:b. In the triangle ABC above, a = 0.6 cm, b = 1.2 cm, whence a/b = 0.7/1.2 = 0.6, rounded to one decimal, that is, tan 31º = 0.6, which means that in a right-angled triangle to any angle = 31º belongs the ratio a/b = 0.6. If you now measure in the figure of the tower, in which =31º, b₁=15 m, b₂=30 m, then once 0.6 = a₁/15m, the other time 0.6 = a₂/30m, whence a₁ = 9 m, a₂ = 18 m.

Comment: In Algebra and Arithmetic, a is a pure number, in Geometry. it is a segment. In trigonometry, both meanings are used. For example, if you write a = 3 cm or a = 3, you refer once to the length of the segment, the other time to the segment itself. There do not arise difficulties out of this usage. The same observation refers to an angle and an area A.

In order to derive quickly the ratio a/b for any angle , you can draw the figure shown: A circle with radius 1 cm and at C its tangent; when, for example,

that is, tan equals the measure of the correponding segment on the tangent, whence comes the term tangent. In order to derive more accurate values, you should draw the figure on millimetre graph paper and use for the smaller angles the radius r = 1 dm, You can thus obtain the table:

Your logarithm table presents more accurate values at smaller intervals.

Applied tasks

During a land survey, you measure angles and segments. Since the magnitude of an angle is more readily determined than the length of longer segments, one prefers for such work measurements of angles and minimizes the number of measurements of lengths.

You can find approximately the magnitude of an angle by means of a protractor, supplied with a plumb line. Better results are obtained with a theodolite, which you know already from geometry.

Adjustment of the screws A allows you to set up the instrument horizontally with the aid of a spirit level . You focus through the telescope, which has inside lines - marked as shown at the top on the left hand side of the figure - in order to make a sharper adjustment possible. Since the instrument can rotate about horizontal and vertical axes, it can be given any orientation.

Depending on whether the point sighted is below or above the horizontal AB, you will speak of angles of depression () or elevation () .

Task: Let the height of the Sun be 40º and the length of the shadow of an edge of a house 15 m. How high is the house?

If you point the telescope at the centre of the Sun - you must use a black screen, to guard your eye sight! -, then, assuming the rays of the Sun to be parallel, - the height of the Sun. The tangent table shows tan 40º = 0.84, whence

0.84 = a/15 m or a = 12.6 m,

and the house is 12.6 m high.

Task: You have measured A₁C₁ = b and the angle and the distance AA₁=d. How high is the tree?

In this task, we will use b = 4 m, = 70º and d = 1.5 m. Since AA₁C₁C is a rectangle, AC = b and CC₁ = d. In the triangle ACB, tan = a/b. Since
tan 70º = 2.75 and b = 4 m, you find that 2.75 = a/4 m or a = 2.75·4 m = 11 m, whence the tree's height is a + d = 11 + 1.5 = 12.5 m.

Task: You know that the tower shown above is 20 m high and have measured the elevation of its top = 60º. How large is AB if d = 1.5 m?

tan = CB₁/A₁B₁, whence A₁B₁ = 0.5/tan . tan 60º = 1.73 and CB₁ = CB-B₁B = a-d = 20-1.5 = 18.5 m, Since in the rectangle BB₁A₁A the sides A₁B₁ and AB are equal, you find AB = 10.7 m.

Exercise

Answer

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