Question 604802: Hello
Can someone help with a trigonometry question please.
PQ and QR are the phasors representing the alternating currents in two branches of a circuit. Phasor PQ is 20A and is horizontal. Phasor QR which is joined to the end of PQ to form triangle PQR is 14A and is of an angle of 35 degrees to the horizontal. Determine the resultant phasor PR and the angle it makes with phasor PQ.
Any help would be gratefully received thanks
David
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Hi David --
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I think I understand the problem from your description, so I'm going to try to help. If I have misunderstood the problem, please post it again and one of the tutors will probably help.
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I'm going to say that phasor PQ has its tail (P) at the origin and extends to the right along the positive x-axis for 20 units (20A) to its head (Q). At this point the phasor QR has its tail and it goes up and to the right at 35 degrees to its head at point R. Its length is 14A. If this is correct, make a sketch of it. (A picture is worth a thousand words of explanation.)
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Now on your sketch, go to point R and drop a vertical line down so that it intersects the x-axis at a point we will call point S. The next thing we are going to do is to break phasor QR into a horizontal component QS and a vertical component SR. (Note that phasor QS will have its tail at Q and its head at S, and phasor SR will have its tail at S and its head at R. Also, note that angle QSR is a 90 degree angle so we have formed a right triangle QSR in which angle RQS is the 35 degree angle. Phasor QS is the adjacent side of the 35 degree angle, phasor QR is the hypotenuse, and phasor SR is the opposite side.
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Using the definition of cosine as being the adjacent side divided by the hypotenuse, we can now write:
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Then we can multiply both sides by QR (and transpose or switch sides) to get:
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And we know that QR is 14 units long, so we can substitute 14 for QR to make the equation become:
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Do the multiplication on the right side of this equation and you should get (to three decimal places) that:
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We can do a similar analyses for the length of SR by using the definition of sine (side opposite/hypotenuse = SR/QR) to get:
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Do the multiplication on the right side and you get the answer the length of SR as being:
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Now we can say that we have formed a new phasor PS that is the sum of phasor PQ and phasor QS. It, therefore, is 20A + 11.468A = 31.468A long. And we also have the perpendicular phasor SR that is 8.030A in length. But observe that phasors PS and SR add together to form phasor PR (tail at P and head at R) and we are trying to find the magnitude (or length) of phasor PR and the angle it makes with positive x-axis.
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We can use trigonometry to find the angle that phasor PR makes with the positive x-axis - the angle RPS) and also the magnitude of phasor PR. We know that in triangle PSR the side opposite to the angle RPS is phasor SR and the side adjacent is phasor PS. With this information we can use the definition of tangent as side opposite divided by side adjacent. Therefore for angle RPS we can write:
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Then substitute 8.030 for SR and 31.468 for PS and this equation becomes:
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Doing the division on the right side results in:
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Rounded to three decimal places. So now all we have to do is find the arctan (or if you prefer, tan^(-1)) of 0.255. Using a calculator and rounding the resulting angle to 3 decimal places we find that the angle RPS is 14.306 degrees. That's one of the answers that we were looking for.
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Now we need to find the magnitude of phasor PR. We could use the Pythagorean theorem because triangle RPS is a right triangle with RP as the hypotenuse, PS as one leg, and SR as the other leg. We could write the Pythagorean theorem as:
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Then substituting 31.468 for PS and 8.030 for SR we get:
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Squaring the two terms on the right, adding those results, and taking the square root of the sum results in finding the magnitude of phasor PR as:
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We can find the magnitude of PR by using either the sine or cosine function. We know that the angle that PR makes with the x-axis is 14.306 degrees and that the side opposite this angle (SR) has a length of 8.030. So we can write the sine function as:
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The sine of 14.306 degrees is 0.247. Substitute that into this equation and you get:
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Multiply both sides by PR and then divide both sides by 0.247 and the equation becomes:
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Do the division on the right side and you get the answer for the length of PR as being:
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and this is fairly close to what we got using the Pythagorean theorem. We could likely decrease the disparity by carrying more decimal places for all our calculations, but that makes a lot more typing necessary.
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Any how, to summarize our results phasor PR has a magnitude of around 32.5A and makes an angle with the x-axis of about 14.3 degrees.
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Please check all this work and analysis. You should probably carry more decimal places in all your calculator results. I haven't had my morning coffee jolt yet, so there are no guarantees about any of this work, but hopefully this discussion is enough to give you the idea of how to proceed with getting the answer to this problem.
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