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 thattan We've been told that tan 20º = 100 / 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 / With a little bit of algebra, we can solve to get
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 = A calculator says that sin 20º = 0.34202. It follows that 0.34202 = 100 / which solves to
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 thattan 0.001º = 10 / A calculator displays tan 0.001º = 1.745329252 x 10 (including plenty of digits to avoid cumulative rounding error), so we know that 1.745329252 x 10 which solves to
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 / Our calculators tell us that sin 0.001º = 1.745329252 x 10 so we have 1.745329252 x 10 which solves to
rounded off to the nearest meter -- the exact same result as we got when we solved for |

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 10^{5} 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 measure180º - (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
It follows, with a little algebra and the help of a calculator, that
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
which we can rearrange to
and use a calculator to work out the result as
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 sum18.0 + 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
A calculator tells us that cos 43.2º = 0.7290 We can simplify and solve the previous formula to get
rounded off to the nearest kilometer. The correct choice is B. |