 In this video we provide the solution to question number 23 from the practice final exam for math 1060 in which case we have a traffic light that weighs 40 pounds and is suspended in the air by these two wires you see right here. The tension along the two wires we'll call it T1 and T2. We want to find the magnitude of the tension in both of these wires here so there's AC and AB like again so tension tension two and tension one right there so we need to find those things so how are we going to do this now since the traffic light is not accelerating towards the earth and not moving at all it's in a state of static equilibrium so if we take these three vectors the weight vector plus T1 plus T2 these have to add up to be the zero vector so because of this the sum of the three vectors adds up to be zero that is if we put these arrows together it makes a triangle it makes a polygon here and so the picture would look something like the following where if we keep the directions the same right your weight is doing something like this then we can move the tensions around so this would be T1 and this would be T2 notice how the directions were preserved in all these situations and so how are we going to figure out these angles this is this is one of the tricky parts about this thing so notice that the angle between the vertical and T1 is 60 degrees that tells us by alter interior angles that this would be 60 degrees as well and likewise if we look at what is the angle between T2 and the vertical well if we draw a vertical line over here that would tell us that this right here is 45 degrees like so which then again by the alternative angle theorem this tells us this is 45 degrees and then the last angle if we subtract that from 180 right 180 minus 45 minus 60 degrees we end up with 75 degrees like so this tells us that this angle the third angle of our triangle is going to be 75 degrees this is an important thing to observe here that this triangle is not a right triangle and so we can't use so katoha right triangle trigonometry here we have to use something like law of signs law of cosines to figure out these lengths first the law of cosines is appropriate but we do know the length of the vector w is 40 pounds so we can use law of signs since we have an angle opposite side pair like so one thing I should caution you about many students when they look at a problem like this are very tempted by this right triangle right here this right triangle doesn't do you a lick of good because although we know this is 60 degrees notice here that the tension is not necessarily the same as the length of this wire right here right so the vector quantity is measuring pounds but this wire length is a length in like feet or inches or something like that right so this right triangle doesn't help you out at all we would want to know this triangle right here for which what is this side what is this side we don't know enough about that triangle so that right triangle won't help us out here same thing if we try to look at this triangle or better yet this triangle we don't know enough about that triangle so this is the triangle we want to be using so using the law of signs we can find this length t1 so we see that t1 over its opposite side sine of 45 degrees this will equal 40 over its opposite side which is sine of 75 degrees like so so we get t1 equals 40 times sine of 45 degrees over sine of 75 degrees for which 45 degrees is root two over two you can of course use your calculator to get an approximation here but the exact value of sine of 75 degrees is actually root six plus root two all over four and so simplifying these things here we're going to end up with 80 root two over root two over the square root of six plus square two that's because we have some fractions side of fractions right here notice that if you're dividing by four in the denominator it's actually multiplying by four on the top four divided by four of course it's two so that doubles the 40 to give us 80 so this is an value right here of course we can also we can also simplify this a little bit a little bit more I should say because if we times the top and bottom by the square root of two like so we see that we end up with a 160 over in the denominator you're going to get two square root of three plus one like so now of course two goes into 160 that happens 80 times and so that sits above now the square root of three plus one that kind of looks like a minus there doesn't have the plus if you want a rational denominator you can times top and bottom by its conjugate the square root of three minus one for which in the denominator you then end up with three minus one which is course 280 times square root of three minus one at the top two goes into 80 40 times and so let me summarize what we have here we get that t1 is equal to 40 times the square root of three minus one that's if you want the exact answer an approximate answer of course would be appropriate here throw that in your calculator to see what the approximate value is going to be don't really care about that right now anyways because again either one's acceptable to find t2 we do a similar calculation right t2 over sine of 60 degrees this is equal to 40 over sine of 75 degrees so in the end this is going to be very very very similar right t2 is equal to 40 times sine of 60 which of course is sine of 60 is going to be three over two and then sine of 75 degrees again is 70 excuse me square root of six plus square root of two over four like so so this gives us 80 times the square root of three over the square root of six plus the square root of two and again if you go to if you go and rationalize this thing again not necessary but the approximate answer would be acceptable but you have lots of various forms that are equivalents to this right here the simplest form would be something like 20 times three root two minus the square of six but again you can go through some of the rationalizing algebra if you wanted to approximate answer would be appropriate here as well