Now angle $angle B = 45^circ$ and therefore $angle A = 135^circ$. If we consider the shape as a triangle, to find the gray line, we must implement the law of cosine with $cos 135^circ$. For example: Since the angle $theta$ is in a triangle, it can be calculated with a little trigonometry, namely the law of cosine. The lengths of the sides of the triangle in Figure 12.3.1 are | $ {bf A}|$, $| {bf B}|$ and $| {bf A}-{bf B}|$. Let $ds {bf A}=langle a_1,a_2,a_3rangle$ and $ds {bf B}=langle b_1,b_2,b_3rangle$; then $$eqalign{ |{ bf A}-{bf B}|^2&=| {bf A}|^2+| {bf B}|^2-2| {bf A}|| {bf B}|costhetacr 2| {bf A}|| {bf B}|costheta&=| {bf A}|^2+| {bf B}|^2-| {bf A}-{bf B}|^2cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2-(a_1-b_1)^2-(a_2-b_2)^2-(a_3-b_3)^2cr &=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2cr &qquad-(a_1^2-2a_1b_1+b_1^2) -(a_2^2-2a_2b_2+b_2^ 2)-(a_3^2-2a_3b_3+b_3^2)cr &=2a_1b_1+2a_2b_2+2a_3b_3cr | {bf A}|| {bf B}|costheta&=a_1b_1+a_2b_2+a_3b_3cr costheta&=(a_1b_1+a_2b_2+a_3b_3)/(|{ bf A}|| {bf B}|) cr }$$ A simple arithmetic with the coordinates of $bf A$ and $bf B$ allows us to calculate the cosine of the angle between them. If necessary, we can use Arccosine to get $theta$, but in many problems it turns out that $costheta$ is all we really need. Okay, I found the answer. The reason we use $angle B$ is that it is equal to $180 – angle A$, which, if implied by $cos$, is equal to $-cosangle A$. Also, I thought the tutorial doesn`t use the normal cosine law, but this: In trigonometry, the law of cosine (also known as the cosine formula, cosine rule, or al-Kashi`s theorem [1]) relates the lengths of the sides of a triangle to the cosine of one of its angles. Using notation as shown in Fig. 1, the law of cosine states We represent a point A in the plane by a pair of coordinates x(A) and y(A) and can define a vector associated with a line segment AB consisting of the pair (x(B)-x(A), y(B)-y(A)).
The easiest way to prove this is to use the vector and point product concepts. Acute case. Figure 7a shows a hepton cut into smaller pieces (in two different ways) to prove the law of cosine. The different pieces are Euclid`s elements that paved the way for the discovery of the law of cosine. In the 15th century, Jamshīd al-Kāshī, a Persian mathematician and astronomer, provided the first explicit statement of the law of cosine in a form suitable for triangulation. He provided accurate trigonometric tables and expressed the theorem in a form suitable for modern use. Since the 1990s, the cosine law in France has always been called the Al-Kashi Theorem. [1] [3] [4] Versions similar to the law of cosine for the Euclidean plane also apply to a unit sphere and in a hyperbolic plane.
In spherical geometry, a triangle is defined by three points u, v and w on the unit sphere and the arcs of great circles connecting these points. If these large circles form angles A, B and C with opposite sides a, b, c, then the spherical law of cosine states that the following two relations are valid: Note that the phrase “projection on $bf B$” is somewhat misleading when taken literally; everything that $bf B$ offers is a direction; The length of $bf B$ does not affect the final vector. For example, in Figure 12.3.4, $bf B$ is shorter than the projection vector, but this is perfectly acceptable. Example 12.3.26 Suppose ${bf z}=| {bf x}| {bf y} + | {bf y}| {bf x}$, where $bf x$, $bf y$ and $bf z$ are all nonzero vectors. Prove that $bf z$ halves the angle between $bf x$ and $bf y$. Now the law of cosine is represented by a simple application of Ptolemy`s theorem to the four-sided cyclic ABCD: Example 12.3.2 Find the angle between the vectors ${bf A}=langle 3,3,0rangle$ and ${bf B}=langle 1,0,0rangle$. We calculate $$eqalign{ costheta &= (3cdot1 + 3cdot0 + 0cdot0)/(sqrt{9+9+0}sqrt{1+0+0})cr &= 3/sqrt{18} = 1/sqrt2cr}$$ so $theta=pi/4$. $square$ However, in the tutorial I saw on the Internet, when calculating the sum of vector forces, they use $cos 45^circ$ in the formula instead and say that we have to use the angle between the vectors (in the tutorial, both come from the point $(0,0)$).
But when we generate the shape geometrically, we don`t technically use the angle facing the edge whose value we want to find. Can someone explain why we use $cos 45^circ$ and not $cos 135^circ$? $ds costheta= {bf A}cdot{bf 0}/(|{ bf A}|| {bf 0}|) =(0+0+0)/ (sqrt{a_1^2+a_2^2+a_3^2}sqrt{0^2+0^2+0^2})$, which is not defined. On the other hand, note that since ${bf A}cdot{bf 0}=0$ it initially seems that $costheta$ will be zero, which, as we have seen, means that the vectors are perpendicular; Only when we realize that the denominator is also zero do we get into trouble. One way to “correct” this is to adopt the convention that the zero vector ${bf 0}$ is perpendicular to all vectors; then we can generally say that if ${bf A}cdot{bf B}=0$, $bf A$ and $bf B$ are vertical. $square$ Although the concept of cosine was not yet developed in its time, Euclid`s elements of the 3rd century BC contain an early geometric theorem that almost corresponds to the law of cosine. The cases of blunt triangles and pointed triangles (corresponding to the two cases of negative or positive cosine) are treated separately in theses 12 and 13 of Book 2. Since trigonometric functions and algebra (especially negative numbers) were lacking in Euclid`s time, the statement has a more geometric connotation: the theorem was popularized in the Western world by François Viète in the 16th century. At the beginning of the 19th century, modern algebraic notation made it possible to write the law of cosine in its current symbolic form.
As in Euclidean geometry, the law of cosine can be used to determine angles A, B, C from knowledge of sides a, b, c. Unlike Euclidean geometry, the reverse is also possible with the two non-Euclidean models: angles A, B, C determine sides a, b, c. In hyperbolic geometry, a pair of equations is collectively called the law of hyperbolic cosine. The first is where, in the identity of the point product, $theta$ is the angle between the vectors. This alone solves your problem and proves that you should use forty-five degrees! One could even argue that if mathematicians (should I say physicists?) three centuries ago focused on the elegant expression of the law of cosine in space 3, the development of vector analysis would have been available to humanity at a much earlier stage; see History of vector analysis. In general, the point product of two vectors is the product of the lengths of their line segments multiplied by the cosine of the angle between them. The numerator of the fraction that gives us $costheta$ often appears, so let`s give it a name and a more compact notation: we call it the point product and write it as $${bf A}cdot{bf B} = a_1b_1+a_2b_2+a_3b_3.$$ This is the same symbol we use for ordinary multiplication. But there should never be confusion; You can tell from context whether we “multiply” vectors or numbers.
(We could also use the dot for scalar multiplication: $acdot{bf V}=a{bf V}$; again, what is meant by context is clear.) Blunt fall. Figure 7b cuts a hexagon into smaller pieces in two different ways, providing proof of the law of cosine in the case where the angle is γ blunt. We have In the limit of an infinitesimal angle, the law of cosine degenerates into the arc length formula c = a γ.