Question

The square of the resultant of two forces 4N and 3N exceeds the square of the resultant of two forces by 12 when they are mutually perpendicular. The angle between the vector is

Answer 1

$(\sqrt{4 {}^{2} + 3 {}^{2} + 2 \times 4 \times 3 \cos( \alpha ) } {}^{} ) {}^{2}$
=
$( \sqrt{4 {}^{2} + 3 {}^{2} + 2 \times 4 \times 3 \cos(90) } ) {}^{2} + 12$
$= > 25 + 24 \cos( \alpha ) = ( \sqrt{25} ) {}^{2} + 12$

$= > 24 \cos( \alpha ) = 37 - 25$
$= > \cos( \alpha ) = 12 \div 24 \\ = > \cos( \alpha ) = 1 \div 2$
So therefore, alpha = 60°
and thus the angle between the vectors is 60°.....

Answer 2

Answer:

The angle between the vector is 60°.

Explanation:

Given that,

Resultant of two forces is

$R^2=(4)^2+(3)^2+24\cos\theta$.....(I)

According to question,

The square of the resultant of two forces exceeds the square of the resultant of two forces by 12.

$R^2=(\sqrt{(4)^2+(3)^2})^2+12$

$R^2= 37$

Put the value of R² in the equation (I)

$(4)^2+(3)^2+24\cos\theta=37$

$25+24\cos\theta=37$

$24\cos\theta=37-25$

$\cos\theta=\dfrac{12}{24}$

$\cos\theta=\dfrac{1}{2}$

$\theta= 60^{\circ}$

Hence, The angle between the vector is 60°.