Find the resultant of two force 130N and 110N respectively acting at angle whose tangent is 12/5

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Resultant of two forces acting at a point calculated by triangle law.

Let 130N force acting along side AB and 110n force acting along side BC. Then resultant will work along side AC of triangle ABC.This resultant is proportional to the side AC.

`AC^2=AB^2+BC^2+2.AB.BC.Cos(theta)`

`tan(theta)=12/5`

`sec^2(theta)=1+tan^2(theta)`

`sec(theta)=sqrt(1+tan^2(theta))`

`sce(theta)=13/5`

`cos(theta)=5/13`

`AC=sqrt(130^2+110^2+2xx130xx110xx5/13)`

`AC=200 `

Thus resultant is 200N along AC and inclined to force 130N by `alpha` , where

`cos(alpha)=13/20`

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