$v_B = 3.94 \textrm{ m/s}$

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This is College Physics Answers with Shaun Dychko. Our job was to figure out how long vector A is and how long vector B is in this picture. Now, the sine law is what's going to do this for us the most efficient way. The sine law says that when you take a side length and divide it by the opposite angle, so consider this side length <i>V A</i> and divide by sine of the opposite angle, which is 23 degrees here, that's going to equal any other side length divided by the sine of its opposite angle. We have the side length <i>V total</i>, divided by its opposite angle, which we can figure out because we know two of the interior angles of this triangle. We can take 180 minus the sum of these two to get this angle here. We can solve for <i>V A</i> by multiplying both sides by the sine of theta <i>A</i>. Then we have <i>V A</i> is <i>V</i> total times sine theta <i>V</i> divided by sine theta of the total. I'll label this theta total because it's the angle opposite <i>V total</i>. Theta total is 180 minus the other two angles, and that's 180 minus 23 minus 26 and a half, which is 130.5. The length of <i>V A</i> then is the length of the total which is 6.72 meters per second. Multiply by sine of 23, which is 23 being the angle opposite VA, and divided by sine 130.5 for length of 3.45 meters per second. The same fancy trick can be used for finding the length of <i>V B</i> as well. We can say <i>V B</i> divided by the sine of the angle opposite this side which is here, theta <i>B</i> equals <i>V total</i> over sine theta total. Then we solve for <i>V B</i> by multiplying both sides by sine theta <i>B</i>. We get <i>V B</i> equals 6.72 times sine 26 and a half degrees divided by sine 130.5 degrees, iving a length of 3.94 meters per second.

## Comments

Submitted by richardfranklins on Tue, 03/31/2020 - 08:22

Submitted by ShaunDychko on Tue, 03/31/2020 - 08:57

All the best,

Shaun

In reply to Could you help me find which… by richardfranklins

Submitted by rickylove on Thu, 05/14/2020 - 20:05

Submitted by ShaunDychko on Sat, 05/16/2020 - 16:57

All the best,

Shaun

In reply to I just got this problem… by rickylove