The calculation of such a value as a tangent may be required both during the solution of trigonometric equations and when searching for a response of the geometry task. It is in the second case that a good help can be a graphic image of an angle, the tangent of which must be found on the cellular paper. How to do this - read in this article.
Work with rectangular triangles
Before starting such a value as a tangent, it is necessary to determine the terminology. So the concept of "tangent angle" characterizes the ratio of the opposite category of the category to the adjacent one. That. Work is carried out within a rectangular triangle.
The essence of the algorithm described below is to work with rectangular triangles within directly determining the Tangent.
Task - Determine the tangent ∠aob.
- Set T. B on the OB beam at the site of its passage through the top of the cell.
- From t. B omit perpendicular on the OA beam. The place of intersection mark as T. C.
- The result is rectangular ΔBoc, in which the angle of ∠aob is located (it is obvious that ∠Boc \u003d ∠aob), the tangent of which must be found.
- Based on the definition of Tangent, TG∠AOB \u003d BC / OC. Looking at the drawing, it is easy to notice that the length of the BC category is folded out of three cell diagonals. In this case, the length of the OC category corresponds to the diagonal of the same cell. Consequently, BC \u003d 3OC.
- tG∠AOB \u003d 3OC / OC \u003d 3.
Task - Determine the tangent ∠aob.
The calculation of TG∠aOB will be based on the fact that Tg (η - λ) \u003d (Tgη - Tgλ) / (1 + TGη * TGλ).
- In one of the points of passing, the rays of the OA and OB vertices of the square cells mark T. a, and so b, respectively.
- Lower those perpendicular. As a result, you get 2 rectangular triangles - ΔOMB and Δola.
- "Calculated" ∠AOB is the difference between the angles of ∠aol and ∠bom: ∠aob \u003d ∠aol - ∠Bom.
- tG∠AOB \u003d TG (∠AOL - ∠BOM) \u003d (TG∠AOL - TG∠BOM) / (1 + TG∠AOL * TG∠BOM). That. Finding the desired value is reduced to finding tangents of angles in constructed rectangular triangles.
- tG∠AOL \u003d AL / OL. Turning to the figure noticeably that Al \u003d 2ol. Therefore, TG∠AOL \u003d 2OL / OL \u003d 2.
- tG∠BOM \u003d BM / OM. Turning to the figure it is clear that OM \u003d 6BM. Therefore, TG∠BOM \u003d BM / 6BM \u003d 1/6.
tG∠AOB \u003d (2 - 1/6) / (1 + 2/6) \u003d 11 * 3/6 * 4 \u003d 11/8 ⇒ TG∠AOB \u003d 1,375.
Using the Kosinus theorem
Task - Determine the tangent ∠aob.
- t. A, and so on, install at the passing points of the specified angle through the vertices of squares. Lower those perpendicular. Also, the segment is connected to each other. A, and T. B.
- Your task is to calculate the lengths of the parties received Δaob. To do this, we appeal to the Pythagoreo theorem.
- AO \u003d √ok 2.+ AK 2By setting the length of the side of the cell as a conditional 1, we obtain AO \u003d √9 + 1 \u003d √10.
- OB \u003d √BP. 2.+ Op. 2, since the length of the cell side is 1, we obtain OB \u003d √4 + 1 \u003d √5.
- According to the cosine theorem, AB 2.\u003d AO. 2.+ OB. 2.- 2AO * OB * COS∠AOB ⇒ COS∠AOB \u003d (AO 2.+ OB. 2.- AB 2) / 2AO * OB. Substitting numeric values, we get:
cos∠aob \u003d (10 + 5 - 25) / 2√5√10;
cos∠aob \u003d -10 / 2√5√10;
cos∠aob \u003d -1 / √2.
- Next, we use the main identity of trigonometry: sinβ 2.+ Cosβ. 2.= 1.
sin∠aob \u003d √1-1 / 2 \u003d 1 / √2.
- It is known that tg∠aob \u003d sin∠aob / cos∠aob \u003d -√2 / √2 ⇒ TG∠AOB \u003d -1.
Depending on the angle, the tangent is to find, choose the most suitable, and the main "working" algorithm.
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