- #1

donniemsb_12

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this example is being discussed by our professor in our class ! ( the one that i upload is the illustration made by our prof in our class )

this are the given and the solution :

A=2.0m @ 40'

B=4.0m @127'

SOLUTION :

R^2=A^2+B^2– 2AB Cos93'

R^2=(2.0)^2 + (4.0)^2 -(2)(2)(4)cos93'

R=4.56m

sinθ/4=sin93'/4.56 sinθ=62.2'

11.2 west of north

guys ! please explain to me how did they get the

this are the given and the solution :

A=2.0m @ 40'

B=4.0m @127'

SOLUTION :

R^2=A^2+B^2– 2AB Cos93'

R^2=(2.0)^2 + (4.0)^2 -(2)(2)(4)cos93'

R=4.56m

sinθ/4=sin93'/4.56 sinθ=62.2'

11.2 west of north

guys ! please explain to me how did they get the

**sinθ=62.2' and 11.2 west of north****and the formula**! thanks ! if u answer this i can now make the assignment given by our prof thanks a lot !