$|B + C| \cos(\theta_3) = |B|\cos(\theta_1) + |C|\cos(\theta_2)$?

Question

My textbook presents the following diagram:

It then claims that, according to the diagram, $|B + C| \cos(\theta_3) = |B|\cos(\theta_1) + |C|\cos(\theta_2)$.

I initially thought that this might be an application of the pythagorean theorem, and it might very well be, but, if so, then it's use of the angles $\theta_1$ and $\theta_2$, in relation to the angle $\theta_3$, is something that I don't understand. On the other hand, given the way the triangle is segmented, it might (also) be some application of similar triangles, but, again, I'm unsure of this.

I would greatly appreciate it if people would please take the time to explain this.

Answer 1

Note that, in the diagram, $B+C$ does not represent the length of a line segment.

In the diagram, $B+C$ represents a vector whose length is $|B+C|$. (Similarly, $A,B,C$ are vectors.)

Also, note that $\theta_1$ is an angle between the vector $A$ and the vector $B$, and that $\theta_2$ is an angle between the vector $A$ and the vecor $C$, and that $\theta_3$ is an angle between $A$ and the vector $B+C$.

Using the definition of cosine, we get $$\cos(\theta_3)=\frac{|B|\cos(\theta_1)+|C|\cos(\theta_2)}{|B+C|}$$ from which $$|B+C|\cos(\theta_3)=|B|\cos(\theta_1)+|C|\cos(\theta_2)$$ follows.

Answer 2

It's just using the definition of cosine ("adjacent side divided by hypotenuse") three times. In the lower left, there is a small right triangle with hypotenuse $B$ and angle $\theta_1$; thus $|B|\cos \theta_1$ is the length of the leg adjacent to the angle $\theta_1$, which is labeled in the diagram. Similarly $|C| \cos \theta_2$ is also labeled for you.

Finally, $|B+C| \cos \theta_3$ is the bottom edge of the largest right triangle in the diagram, since that triangle has $B+C$ as the hypotenuse and $\theta_3$ as an angle. Clearly it is the sum of the two smaller segments $|B| \cos \theta_1$ and $|C| \cos \theta_2$.

Answer 3

This is simply the sum of two vectors, $B$ and $C$. By the formula of the sum ($A=B+C$), I have: $$|A_x|=|C|\cos(\theta_2)+|B|\cos(\theta_1)$$ and $$|A_y|=|C|\sin(\theta_2)+|B|\sin(\theta_1)$$ Now, note that by the teorem on the rectangular triangle, I can say: $$|B+C|\cos(\theta_3)=|C|\cos(\theta_2)+|B|\cos(\theta_1)$$