Problem

Forces of \(40~N\) and \(55~N\) act with an included angle of \(70°\), as shown. Find the resultant magnitude and its direction measured from the \(40\text{-}N\) force.

Prompt vector diagram with forces 40 N and 55 N separated by 70°; resultant magnitude and direction from the 40-N force are withheld.

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What this problem is really about

Use the first vector as a reference axis and resolve the second into parallel and perpendicular components. Add the parallel contribution to the first magnitude while retaining the perpendicular contribution; those two totals determine the resultant's magnitude and direction. For tail-to-tail addition, the equivalent cosine expression carries a positive cross term.

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Use the correct magnitude 78.3 N but report 70°-41.3°=28.7° as though it were measured from the 40-N force.

R=√(40²+55²+2·40·55cos70°)=78.3 N, and direction=atan[(55sin70°)/(40+55cos70°)]=41.3° from the 40-N force.

Use a minus sign: R=√(40²+55²-2·40·55cos70°)=55.9 N, while still reporting 41.3°.

Add magnitudes and ignore the angle, reporting R=95 N at 0° from the 40-N force.

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Forces of \(40~N\) and \(55~N\) act with an included angle of \(70°\), as shown. Find the resultant magnitude and its direction measured from the \(40\text{-}N\) force.

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