Question

Which of the following cannot be the angles made by a force with the co-ordinate axes

Answer 1

Answer:

100,40 and 3 is a answer

I hope my answer is correct

Answer 2

Answer:

100 and 40 cannot be the angles made by a force with the co-ordinate axes

Explanation:

It is given that,

|A| = |B|

Using the law of cosine to calculate the magnitude of the resultant vector R.

$|R|^2 = |A|^2 + |B|^2 + 2|A||B| . cos (\theta)\\|R| = \sqrt( |A|^2 + |B|^2 + 2|A||B| . cos (\theta))\\|R| = \sqrt|A|^2 + |A|^2 + 2|A||A| . cos (\theta)$

$|R| = √ ( 2|A|^2 + 2|A|^2 . cos (\theta) )\\|R| = √ ( 2|A|^2 ( 1 + cos (\theta)) )$

Applying half angle identity,

$|R| = \sqrt (4A^2 cos^2 ( \theta / 2))\\|R| = 2 A cos ( theta / 2 )$

Now, for calculating the resultant angle α that it will make with the first vector,

$tan \alpha = ( A sin \theta ) / ( A + A cos \theta )\\tan \alpha = (2 A cos (\theta / 2) . sin (\theta / 2) / ( 2 A cos2 (\theta / 2))\\tan \alpha = tan (\theta / 2)\\ \alpha = \theta / 2$

Hence, this shows that the resultant will bisect the angle between the two vectors having equal magnitude.

Let’s consider two vectors, A  and  B, and the resultant of two vectors is R.

Hence, according to the condition  given in the question:

$|A| = |B| = |R|$

Now, according to the law of cosines,

$|R|^2 = |A|^2 + |B|^2 + 2|A||B| . cos (\theta)$

Since, |A| = |B| = |R|

$|A|^2 = |A|^2 + |A|^2 + 2|A||A| . cos (\theta)\\|A|^2 = |A|^2 + |A|^2 +  |A|^2 . cos (\theta)\\|A|^2 = 2|A|^2 + |A|^2 . cos (\theta)\\|A|^2 = 2|A|^2 ( 1 + cos (\theta) )\\|A|^2 / 2|A|^2 = ( 1 + cos (\theta) )\\1 / 2 = 1 + cos (\theta)$

So, the angle between two vectors having equal magnitude is equal to 120º.