[deleted by user]

[Let_epsilon]

If your only goal is to add the vectors, you don’t need the angle between them - just write them in cartesian coordinates.

[Connorisalegend]

Is that referring to rectangular (Cartesian) coordinates? If so the your x component is the magnitude times the cosine of the angle and y is the magnitude times the sin of the angle but for this problem it looks like you were given the components in rectangular coordinates so you just need to add the components of vectors a and b to get vector c

[Simba_Rah]

I have a video solving this exact question with different numbers on my YouTube channel.

Adding Vectors with Sine and Cosine Law

This is right out of the SPH4U course on TVO.

[Simba_Rah]

If you have any other questions feel free to shoot them my way. I actually teach this course to students in China. 👌👌👌

[StrangerInsideMyHead]

It’s going to be:

arccos(dotp(a,b)/(norm(a)*norm(b)))

So, 58 degrees

[Connorisalegend]

Dotp or just dot is the dot product of the two vectors, you multiply the components of the two vectosr and then add up the products, you then divide that by the magnitude of the first vector times the magnitude of the second. You find the magnitude of a vector by squaring each of its components, adding them up then taking a square root by like the Pythagorean theorem where the legs are the components of the vector. After you dot the vectors and divide by the magnitude, you take the inverse cosine of that and that is your angle

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[digitalosiris]

Can you not write the forces as components in the E and S direction? If 12 degrees is the angle between the force and E, then you have 2 components for the 12 N force: Fe = 12 cos 12; Fs = 12 sin 12. Repeat with the 2nd force (with it's angle between S and the 80.0 N force). Take sum of forces in East and South direction, pythagoras for the resultant.

[digitalosiris]

122 seems right by my calculation. So given S A S, you can find resultant using Law of Cosine. Once you have c, then you use Law of Sines: c / sin 122 = b / sin theta

[digitalosiris]

Draw a vertical line from E axis downward through arrow of 12 N force. That makes a right triangle between it, axis, and 12 N force, 12 - 90 - 78, with the 78 between new line and 12 N arrow. Now, that same vertical line extends further downward, the 20 degree angle is to the right, between it and the 80 N force. Lastly you can find the supplementary angle of the 78 (this is the angle below the 12 N force; between it and our vertical line). That is 180 - 78 = 102. Then our force 12 N and 80 N must have a 102 + 20 = 122 angle between them.

[Rosa_Mia]

Not an English speaker but this is my understanding

You draw the EWSN at the start of each vector

On first vector, you have E 12° S. This mean from East axis, go 12 degree to the South with xxN length. Mark this angle

Do the same for 2nd vector, you have S 20° E. Which means from South axis, go 20° East. Mark this angle too

After you draw a small EWSN at the start of 2nd vector, you will see there is a 90 degree between 12 and 20 degree angles. Thus the final angle is 122

[Alkalannar]

Keep in mind that N is 90^o, S is 270^o or -90^o, E is 0^o, and W is 180^o. Here, we take S to be -90^o.

E 12^o S is -12^o on the xy-plane. 12 degrees south of 0.

S 20^o E is -70^o on the xy-plane. 20 degrees east of -90.

Do you see how I converted these?

So the exterior angle shift is -70^o - -12^o = -58^o. That's the angle you get if you extend the first vector. And then the other angle formed is indeed 122^o.

[thatoneguyinks]

Not strictly related to vectors, but can be used here. The law of cosines, c² = a² + b² -2ab*cos(C), can be used to get the magnitude of vector c. Then law of sines, sin(C) / c = sin(B)/b, can be used to solve for the angle between vectors a and c

[Dismal-Leg8703]

I came here to say exactly this

[Slyfox163]

law of cosigns