| Trigonometry Demystified, 2nd edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 9 |
| 1. In Fig. 9-13, the measure of θ equals the measure of ∠QRP
at the far right-hand end of the right, because they're alternate interior angles to the
transversal QR that crosses a pair of parallel lines. Based on that information and
according to the right-triangle paradigm, we know that tan θ = b / x We've been told that θ = 20.0º and b = 100 m. When we input these values to the above formula, we get tan 20º = 100 / x We can use a calculator to find tan 20º = 0.36397 (going out to a couple of extra digits to prevent cumultive rounding error), so 0.36397 = 100 / x With a little bit of algebra, we can solve to get x = 275 m rounded off to the nearest meter. The correct choice is C. |
| 2. Using the same geometry as we applied when we solved the previous problem, we see
that sin 20º = b / y A calculator says that sin 20º = 0.34202. It follows that 0.34202 = 100 / y which solves to y = 292 m rounded off to the nearest meter. The answer is B. |
| 3. We're told that b = 10 m (exactly) and also that θ = 0.001º
(exactly), so we don't have to worry about significant figures when we solve for x.
We're still working with Fig. 9-13, so the right-triangle paradigm tells us that tan 0.001º = 10 / x A calculator displays tan 0.001º = 1.745329252 x 10-5 (including plenty of digits to avoid cumulative rounding error), so we know that 1.745329252 x 10-5 = 10 / x which solves to x = 572,958 m rounded off to the nearest meter. The correct choice is A. |
| 4. Following the geometry of Fig. 9-13 with the new information, we have sin 0.001º = 10 / y Our calculators tell us that sin 0.001º = 1.745329252 x 10-5 so we have 1.745329252 x 10-5 = 10 / x which solves to x = 572,958 m rounded off to the nearest meter -- the exact same result as we got when we solved for x! (Actually, y exceeds x by a small distance, but evidently it's less than a meter, despite the large values of both x and y.) The correct choice, once again, is A. |
| 5. This question is a little tricky (but not too bad, if we pay reasonable attention!). We need to remember that the parsec (pc) represents a fixed unit of distance in space, even though the acronym technically stands for "parallax second" which suggests an angle. We've been told that the more distant star lies 10 times as far away from us as the nearer star does (470 pc as compared to 47 pc). On an interstellar scale, the amount of parallax that we observe varies inversely with the distance. Therefore, we'll observe only 1/10 as much parallax for the star at 470 pc as we see for the star at 47 pc. The correct choice is B. |
| 6. The parsec (pc) constitutes a fixed unit of distance equivalent to about 3.262 light years or 2.063 x 105 astronomical units (AU). A star 470 pc away is therefore 10 times as distant as a star 47 pc away. The correct choice is C. |
| 7. Given a triangle that lies on a flat surface, if we divide the length of any side by the sine of the angle opposite that side, we always get the same ratio. None of the choices here say anything of that sort. The answer is therefore D, "None of the above." |
| 8. We can use the law of sines to find the length of side x in the triangle
of Fig. 9-15. We know that one of the sides is 40 m long. The figure doesn't tell us the
measure of the angle opposite that side, but we can calculate it as 180º minus the sum of
the measures of the other two angles. Basic geometry reminds us that the measures of the
interior angles of a plane triangle always add up to 180º, so the angle opposite the 40-m
side must measure 180º - (95º + 40º) = 180º - 135º Now let's find the triangle's "characteristic ratio" 40 / (sin 45º) = 40 / 0.7071 We can go to more than two significant figures here, because we've been told that all three of the given values (two angle measures and one side length) are exact. Four significant figures should be enough to avoid cumulative rounding errors. According to the law of sines, this "characteristic ratio" for the triangle, 56.57, is the ratio of the length of side x to the sine of the angle opposite it, which mesures 40º. In mathematical terms, we have x / (sin 40º) = 56.57 It follows, with a little algebra and the help of a calculator, that x = 56.57 sin 40º rounded off to the nearest meter. The correct choice is D. |
| 9. In the scenario of Fig. 9-15, the "characteristic ratio" for the triangle
(the length of any side divided by the sine of the angle opposite that side) is 56.57; we
worked that out when we solved the previous problem. The law of sines therefore assures us
that y / (sin 95º) = 56.57 which we can rearrange to y = 56.57 sin 95º and use a calculator to work out the result as y = 56.57 x 0.9962 rounded off to the nearest meter. The answer is B. |
| 10. Warning: This solution gets a little tedious! Let's start by figuring out
the measure of the angle between the beacons' azimuth bearings as seen by an observer
on the vessel at the center of Fig. 9-16. We're told that beacon X lies at
azimuth 342.0º, which is 18.0º west of due north. We're also told that beacon Y
lies at azimuth 25.2º, which is 25.2º east of due north. The angle between the two
beacons, as seen from the vessel, therefore equals the sum 18.0 + 25.2º = 43.2º We've been given the lengths of two sides of a triangle whose vertices correspond to the vessel's location and the beacon locations. Those two sides lie along the dashed lines in Fig. 9-16. The distance from the vessel to beacon X equals 38.1 km, and the distance from the vessel to beacon Y equals 44.8 km. Now we know the lengths of two sides of the triangle. We also know the measure of the angle between those two sides. The law of cosines tells us that we can solve for the distance between the beacons (let's call it d) using the rather messy formula d = [38.12 + 44.82 - (2 x 38.1 x 44.8 x cos 43.2º)]1/2 A calculator tells us that cos 43.2º = 0.7290 We can simplify and solve the previous formula to get d = [1452 + 2007 - (2 x 38.1 x 44.8 x 0.7290)]1/2 rounded off to the nearest kilometer. The correct choice is B. |