p and q are two components of a given force f and its line of action divides the angle between them in the ratio of 1:2 ,prove that q(f+q)=p^2

Answer 1

Unfortunately I am not able to attach a picture, so please try to draw the following:

Draw a triangle with angles α, 2α, and 180°- 3α, for some arbitrary α.

Let

p be the length of the side facing angle 2α,

q be the length of the side facing angle α,

f be the length of the side facing angle 180°- 3α.

Should you draw the situation given in your problem with forces p and q with vector sum f,

you would get the typical parallelogram.

The force f would divide the parallelogram into two triangles, both congruent

with the triangle I asked you to draw.

So from now on we will deal only with that triangle.

We use the law of sines and some trigonometric identities

p/q = sin(2α)/sin(α) because Law of sines

= 2sin(α)cos(α)/sin(α) because sin(2x) = 2sin(x)cos(x)

= 2cos(α)

f/q = sin(180°- 3α)/sin(α) because Law of sines

= sin(3α)/sin(α) because sin(180°- x) = sin(x)

= (3sin(α)cos²(α) - sin³(α))/sin(α) because sin(3x) = 3sin(x)cos²(x) - sin³(x)

= 3cos²(α) - sin²(α)

= 3cos²(α) - (1 - cos²(α)) because sin²(x) = 1 - cos²(x)

= 4cos²(α) - 1

= (2cos(α))² - 1

= (p/q)² - 1

You should be able to manipulate the derived identity

f/q = (p/q)² - 1

into the desired form

q(f+q) = p²