Solutions for practice questions: Chapter 7, Triangle Trigonometry If you find any errors, please let me know at

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1 Solutions for practice questions: Chapter 7, Triangle Trigonometry If you find any errors, please let me know at The shortest distance from the chord [AB] to the center O of the circle shown below is units, and the circle s radius is 5 units. We are to find ˆ sin AOB. To make the diagram more useful, I ll draw in the radii from O to the endpoints of the chord: 5 5 The distance from a point to a line is measured along the perpendicular, so I ll draw that in, too. 5 5 Hey, right triangles! Sadly, the angle I m looking for the sine of is not one of those in either of the right triangles; it s AOB. Since this isn t in one of the right triangles, the sine ratio of opposite over hypotenuse won t be any help. I will need the law of sines. In AOB, that sets up to look like sin AOB ˆ sinoab ˆ sinoba ˆ. While AB OB OA I also don t know the other two angles, I can find either one of their sines quite easily they are in right triangles. ˆ opp sinoab hyp 5 sin AOB ˆ So far, this gives me 5. If AB 5 only I knew the value of AB. But look! The little triangles are --5 right triangles. So AB is two of those fourunit-long legs, and AB = 8. Then sin AOB ˆ 5. Multiplying by 8 gives 8 5 ˆ 5 sin AOB In a right triangle, tan =, and then 7 we re supposed to find sin and cos. Why would they bother to tell us it s a right triangle? Well, the double-angle formulas have sine and cosine in them. If we know the tangent of an angle in a right triangle, that gives us the opposite and adjacent sides, and we can find the hypotenuse. hyp 7 58 That will let me find the sine and opp cosine of : sin, and hyp 58 adj 7 cos. hyp 58 You know, Pythagorean triples.

2 Then sin = sin cos = , and cos = sin = Pretty The three sides of a triangle are, 5, and 7 units long. Find the largest angle. Using a theorem from geometry, the largest angle will be across from the largest side. That means we re looking for the angle across from 7. I ll draw a picture. So the largest angle will be B. With three sides known, I ll use the law of cosines. 5 7 cos B, and mb cos Angle A is obtuse, and its sine is 5. Find sin A exactly. The exact value part of the question is to keep you from doing calculator tricks. The formula for sin A is sin A cos A, so I should start by finding the cosine of the angle. This can be done with a triangle, like I did earlier, or with a I tried to find the name of this theorem, but I ve come to the conclusion that it probably doesn t have one. Sorry. formula. This time, I m going for the formula. sin + cos = 5 cos A 5 cos A 69 cos A 69 cos A And here s where the obtuse part comes in. Obtuse is second quadrant, and in QII, cosine is negative. So cos A. Finally, 5 0 sin A In Given the diagram at right, the distance BQ is given as 0 m, and the angles are as labeled. Notice that 0m this is a threedimensional picture; B, A, and Q are all on the same level ground, and P is the top of a pole stuck in the ground at Q. I have labeled the 0 m side. Having tried to label that side clearly, I now understand why BoB didn t do that himself. It s awkward. There are four different triangles in the diagram. I m sure we re going to need more than one of them. a) To find PQ, I ll definitely need to use PQB, as BQ is the only length I know. The angles alone can tell the shape of the triangles, but it takes a side to get the size. So here s a diagram of that one triangle alone:

3 6. The given diagram looks sadly incomplete. I m going to finish the triangle. The required length, PQ is opposite the 6 angle, and we know the adjacent side. It s known to be a right triangle, so this looks like a case for the tangent ratio. PQ tan 6 0 PQ 0 tan 6 9. m b) The distance between A and B will also need to be part of a triangle in which we know a side. In this case, I m thinking it s the one on the ground, ABQ. A picture: In case you re wondering, the diagrams that look like this one are done with GeoGebra. This one and the one in part (a) are to scale. This one is not a right triangle; the angle at Q is 80. We need to find AB, and with the value of mbqa, the law of sines is a good option. AB 0 sin80 sin 70 0sin 80 AB.9 m sin 70 The proof is left as an exercise for the reader. Don t tell me you ve already forgotten. 56 km To find the measure of CAB, I ll be using the law of cosines. ˆ 8 56 cos CAB 8 ˆ 8 56 mcab cos For this one, the diagram sort of seems like overkill, at least in part (a), A but I ve included it for your viewing enjoyment. a) The smallest angle will be opposite the smallest side. I ve labeled the vertices to make it easier to talk about the sides and angles. Since L is opposite the smallest side, that s the one we need to find. This will take the law of cosines. Again cos L mlcos (The question says in degrees, but does not specify a precision, so I did s. f.) b) For the area of the triangle, I ll use the formula A absin C. That requires two sides and an included angle. In this case, the C L

4 only angle I know is L. 5 The two sides we need will be CL and AL, because they are the two segments that form L. The area is therefore CL AL sin L 78sin8. 7. cm. 8. I have labeled the information given in the diagram to make it easier to digest. 0 cm 7 cm 0 cm cm a) The reason they ve made two diagrams is so that you can see triangle is obtuse. 6 For ACB, we ll need the law of sines, and we are looking for the aforementioned obtuse angle. sin ACB ˆ sin ˆ 0sin 50 macb arcsin Wait! That s not obtuse!!! This is that ambiguous case you know and love. There are two angles with the same value of sine, and the other one is this one s supplement. So while that was the measure of ACB, it was in the wrong triangle. Really, the 5 I had initially called that one C in the diagram, but I ve just gone back and changed it so that it doesn t look like you can only use angles with that name! 6 Yes, in a problem like this, you are allowed to assume some to scale for comparison s sake. Yes, it would have been helpful if they d said that explicitly. one we want is macb (to three significant figures). b) To find the area of triangle, we need two sides and an included angle. The two sides lengths we have are AB and AC, so we ll need mbac. I can get that by subtraction. mbac Then the area we need is 0 7 sin cm. Note that throughout the problem, even though I ve typed the answers to s. f., I ve kept more digits in the calculator and used them to do the computations. 9. Boats A and B move away from point P at a 70 angle from each other with speeds of 0 km/h and km/h, respectively. How far apart are they after.5 hours? This calls for a diagram. After.5 hours, boat A will have traveled.5 0 = 50 km, and boat B will have gone.5 = 80 km. Those are the distances marked in the picture. The distance between the books is AB, and it can be found with the law of cosines. AB = AP + BP AP BP cos 70 = (50)(80) cos , so AB 78.5 km.

5 0. Find JKL in the diagram.. Oh, this looks awesome! The angle we need is opposite the 5 unit side, so the law of sines should work. sin JKL ˆ sin ˆ 5sin 5 sin JKL 8 ˆ 5sin 5 mjkl sin (to 8 the nearest degree). Here s the diagram. It would be nice to have a model that you could turn around and see from various angles. A rotating D model is way too much hassle to try to embed in this document, but I ve made a diagram that shows the relevant rectangle head-on. The sides of the rectangle are in red superimposed on the picture from the book. The rotated version is beneath it. a) The length AB is opposite the 0 angle. If we knew the angle opposite BC, we could use the law of sines. Since the sum of the angles in a triangle is 80, mbac ˆ AB 5 sin 0 sin80 5sin 0 AB.6 cm sin80 b) Since we now have sides AB and BC, the area is AB BC sin ABC ˆ.6 (5)sin cm. We need some side lengths. All cubes are symmetric, so a side length of for the side of the cube (the height of the red rectangle) would be fine. We ll also need the base of the red rectangle, which is a diagonal of the square base of the cube.

6 Clearly the length of the red diagonal can be found using the Pythagorean theorem to be. 7 Then the measurement of the base of that red rectangle above is. Remember, we re looking for the angle between the diagonals. While all of the angles of the rectangle are right angles, that doesn t give any of the angles of the four small triangles. If we knew the diagonal lengths, though, they are bisected where they intersect. Let d be the length of a diagonal. Then, so d. d Each black side of the small triangle is half that long. One last diagram. (This problem is much easier to work than it is to type up.) And finally all we have to do is find the measure of the angle in green there. I m going to call it. This will take the law of cosines. cos cos Done.. So there s a crazy triangular field. Initially, we only need part of the information, so I m going to make a diagram with just the initial measurements. a) Using the law of cosines, BC cos60 = 88, and BC = 9 m. b) Area can be found with 650 sin = 690 m. Now there s a point D on [BC] so that [AD] bisects the 60 angle. The length of [AD] is called x. It s time for the diagram from the book. 7 Please don t tell me you don t understand why. Please.

7 c) i) The area A can be found as A 65 x sin0 = 65x 65 x. Q.E.D. ii) A 0 x sin0 = 0x 0 x 6x iii) So how to use this information to find the value of x? It goes back to what we found in part (b). The total area is 690 m, and we know the two pieces in terms of x. So the equation to solve is 65x 6x x 690 x m. 69 d) i) In the diagram, ADC and ADB are supplementary. The short reason, then, is that the sines of supplementary angles are equal. The longer one is that supplementary angles have the same reference angle; one is in the first quadrant and the other is in the second quadrant, and in ii) both cases, the sine is positive. So sin ADC ˆ sin ADB ˆ. The law of sines only applies to a single triangle, and BD and DC are not sides of a single triangle. That makes this a nontrivial question. However, it follows part (i), and that s a hint. The sines of the two angles at D are the same. So I ll use the law of sines of each of the smaller triangles. In ADC, DC 0. In sin 0 sin ADC ˆ BD 65 ADB,. sin 0 sin ADB ˆ Rearranging a little so that the angles are all on one side of each equation, the first one becomes DC sin0, and the 0 sin ADC ˆ second becomes BD sin 0. Because 65 sin ADB ˆ sin ADC ˆ sin ADB ˆ, the right sides are the same, and the transitive property of equality leads to DC BD. We can 0 65 exchange the extremes of the proportion 8 to get 65 BD DC 8 8 Q. E. D. 8 As long as the cross-products are equal, the proportion will still be true.