Video Transcript
Find the measure of the acute angle that lies between the straight line whose direction vector is 𝐫 is equal to one, negative three and the straight line whose equation is negative two 𝑥 minus five 𝑦 plus one is equal to zero in degrees, minutes, and the nearest second.
In this question, we’re asked to find the measure of the acute angle between two straight lines. We need to give our answer in degrees, minutes, and to the nearest second. We can do this by recalling the formula for finding the measure of this acute angle in degrees. We can recall if we have two straight lines of slopes 𝑚 sub one and 𝑚 sub two, then the acute angle 𝛼 between the two straight lines will satisfy the equation the tan of 𝛼 is equal to the absolute value of 𝑚 sub one minus 𝑚 sub two divided by one plus 𝑚 sub one times 𝑚 sub two.
And there’s a few things worth noting about this formula. First, if our two straight lines are parallel, then 𝑚 sub one will be equal to 𝑚 sub two. Next, if the two straight lines are perpendicular, one of two things can happen. Either 𝑚 sub one and 𝑚 sub two will multiply to give negative one, in which case we’ll have a zero in the denominator of this expression. So, tan of 𝛼 is undefined, which means 𝛼 is 90 degrees. Or one of the lines is vertical, which means it has an undefined slope. So, we’d need to check this case separately. However, as we’ll see, we don’t need to worry about the edge cases in this question.
So, let’s start by finding the slopes of the two lines given to us in the question. Let’s start with the first line, the line with direction vector 𝐫 one, negative three. We can determine the slope of this line directly from its direction vector. We can recall if a line has a direction vector 𝐝 equal to 𝑎, 𝑏, then the slope of the line 𝑚 will be equal to 𝑏 divided by 𝑎. And this is provided 𝑎 is nonzero. If 𝑎 is equal to zero, then our line will be vertical. We can apply this to find the slope of the line. 𝑚 sub one will be equal to negative three over one, which is equal to negative three.
It’s also worth noting we can see this directly from the direction vector itself. We note the direction vector runs parallel to the line. And we can see for every one unit we move to the right, we move three units down. So, the change in 𝑦 over the change in 𝑥 is negative three over one, which is negative three.
We now want to determine the slope of our other line. We can see this is given in general form. And it’s not easy to see the slope of a line given in general form. So, instead, let’s rewrite this into slope–intercept form. We do this by first adding two 𝑥 to both sides of the equation and subtracting one from both sides of the equation. This gives us negative five 𝑦 is equal to two 𝑥 minus one. Then, we’ll divide the equation through by negative five and simplify. We get that 𝑦 is equal to negative two-fifths 𝑥 plus one-fifth. And finally, we know in the slope–intercept form, the coefficient of 𝑥 will be the slope. So, 𝑚 sub two will be negative two over five.
We can now substitute these values for the slopes into our formula. This gives us the tan of 𝛼 is equal to the absolute value of negative three minus negative two-fifths divided by one plus negative three times negative two-fifths. And now we can start evaluating this expression. First, in our numerator, negative three minus negative two-fifths is negative three plus two-fifths. We can calculate this; it’s negative 13 over five. Next, in the denominator, we have one plus negative three times negative two-fifths. Negative three times negative two-fifths is six over five, and we add one to get 11 over five.
Therefore, the tan of 𝛼 is the absolute value of negative 13 over five divided by 11 over five. And we can simplify this equation. We can start by canceling the shared factor of one-fifth in the numerator and denominator. Next, we can see we’re taking the absolute value of a negative. So, we can just remove the negative symbol. This gives us 13 divided by 11. Therefore, we’ve shown the tan of 𝛼 is equal to 13 divided by 11. We can solve for 𝛼 by taking the inverse tangent of both sides of the equation. We’ll then evaluate this by clearing some space and then checking that our calculator is set to degrees mode. 𝛼 is the inverse tan of 13 over 11, which is equal to 49.763 and this expansion continues degrees.
But remember, we’re told in the question we need to give our answer in degrees, minutes, and to the nearest second. However, this angle is only measured in degrees. So, we’re going to need to convert this angle into degrees, minutes, and seconds. To do this, we’ll start by recalling there are 60 minutes in a degree and 60 seconds in a minute. We can then use this to convert our angle. We’ll start by noting there are 49 degrees in our angle 𝛼. This means we need to convert the remaining expansion into degrees and minutes. Since this is 0.763 and this expansion continues degrees and there are 60 minutes in a degree, we’re going to need to multiply this value by 60.
And when we do this calculation, it’s very important that we use the exact value for our angle in degrees. We can do this either by using the memory function in our calculator or by using an exact expression for this angle, the inverse tan of 13 divided by 11 minus 49. In either case, multiplying 0.763 and this expansion continues degrees by 60 gives us 45.818 and this expansion continues minutes. So, we have 45 minutes.
We now need to determine the number of remaining seconds in this angle. And we can do this by noting we have 0.818 and this expansion continues minutes, and there are 60 seconds in a minute. So, once again, we multiply the remaining angle in minutes by 60. This will convert the angle into seconds, where we use the exact value of this angle. This gives us 49.110 and this expansion continues seconds. And we’re told in the question we need to give our answer to the nearest second. So, we need to round this value. We see the first decimal place is one, so we round this value down. And this gives us our final answer.
The measure of the acute angle between the line with direction vector one, negative three and the line with equation negative two 𝑥 minus five 𝑦 plus one is equal to zero in degrees, minutes, and seconds to the nearest second is 49 degrees, 45 minutes, and 49 seconds.
