Question

If the resolved part of a force in the horizontal and vertical directions are equal in magnitude, the angle at which it is inclined with the horizontal is

Answer 1

Let the force be x

Hence

Horizontal component=xcos(theta)

Vertical component=xsin(theta)

According to given condition

Horizontal component=Vertical component

xcos(theta)=xsin(theta)

hence

cos(theta)=sin(theta)

This is possible when

Theta=45°

Hence the angle at which it is inclined is 45°

Hope it helps

:)

Answer 2

Answer:

45°

Explanation:

we know that the components of a force is[ sin theta ] and [ cos theta ]

so if the force is "F" then vertical component is Fsin(theta) and horizontal component is Fcos(theta)

so , as per question we know

vertical component = horizontal component.

so, Fsin(theta)=Fcos(theta)

or, sin(theta)=cos(theta)

or, sin(theta)=sin(90°-theta)

or, theta=sin(Inverse)[sin(90°-theta)]

or, theta=90°-theta

or, 2theta=90°

or,theta=45°