; M.METKY. 



in 



oaae may be treated. It will be found that other ratios might 

 be taken equally well in several instances. 

 EXERCISE 2. 



1. If a - 30-660 and b - 17, find c, A, and B. 



2. If e m dM) and B - 29, flnd a, b, and A. 



3. If I m 4-5 and B 54, flnd , o, and A. 



4. If A - 61 and b - 22, flnd a, e, and B. 



5. If a - 670 feet and b = 833 yards 1 foot, find e, A, and B. 



6. If a = 17<U and c = 3000, flnd b', A, and B. 



7. If A = 75 and o ^ -005, flnd a, b, and B. 



8. If b - -875 and e = 1, find A, B, and a. 



9. If o - 120 and a = 77$, find A and B. 



10. A house 50 feet high abuts upon a street found to measure 33'7 

 feet in width. Find the length of ladder required to reach the top 

 from the opposite side of the street, and the angle the ladder will make 

 with the wall of the house. 



11. Two trains travelling, one at 20 miles an hour, the other faster, 

 come into collision at a level crossing, where the two lines (both being 

 free from curves) cross each other at on angle of 36. Some time before 

 the collision, a passenger in the slower train observes the other exactly 

 abreast of him on the other line of railway, and judges the trains to be 

 ft quarter of a mile apart. How far from the crossing were both trains at 

 that moment, and what was the speed of the faster train ? 



12. One of two boys, flying a kite in a level field, observes that he has 

 let out the whole of bis string 60 yards just as his companion, look- 

 ing up, cries out that the kite flies perpendicularly over his head. The 

 boys found afterwards they were standing 30 yards apart. How high 

 was the kite, and what angle did the string make with the ground ? 



13. The rope holding the "captive balloon" at Chelsea against a 

 strong wind, when 400 yards were paid out, was found to incline 15 

 from the perpendicular. How high was the balloon, and how far from 

 the foot of the rope would a piece of iron ballast have fallen, if dropped 

 from the balloon P 



XI. The Fundamental Formulae. We have hitherto examined 

 only the relations between ratios of the same angle ; we pro- 

 ceed now to trace the relations between ratios of two or more 

 different angles. The number of formulae expressing these 

 relations may be extended almost at will, but they are all de- 

 rived from the following formulae for the sines and cosines of 

 the sum and difference of two angles, known, therefore, as the 

 four fui> iamental formula . 



Sin. (A + B) = sin. A cos. B -f cos. A sin. B (33) 



Sin. (A - B) = sin. A cos. B cos. A sin. B (34) 



Cos. (A + B) = cos. A cos. B sin. A sin. B (35) 



Cos. (A - B) = cos. A cos. B + sin. A sin. B (36) 



where A and B are any angles whatever. 



These formulae may be thus expressed in words : 



(33) The sine of the sum of two angles is equal to the sine of 

 the first into the cosine of the second plus the cosine of the first 

 into the sine of the second. 



(34) The sine of the difference of two angles is equal to the 

 sine of the first into the cosine of the second minus the cosine of 

 the first into the sine of the second. 



(35) The cosine of the sum of two angles is equal to the pro- 

 duct of their cosines minus the product of their sines. 



(36) Tlie cosine of the difference of two angles is equal to the 

 product of their cosines plus the product of their sines. 



To prove (33). In Fig. 8 let A o B = A, and B o c = B ; then 

 A o C = A + B. In this case 

 B the sum of the angles exceeds 

 one right angle, but the same 

 construction and lettering hold 

 good if the sum be taken as 

 less than 90*, though the 

 figure will be differently ar- 

 A ranged. 



_ ~O s In o c take any point, P, and 



from it draw p Q, p R, perpen- 

 dicular to A o (produced) and to B o. From R draw R T and R s 

 perpendicular to p Q and A o. 



_PQ_QT + PT 



Fig. 8. 



/A-LM\ * o,o rm 

 .'. mn. (A + B) = - . + 



OB OP OB OP 



= sin. A co. B + con. A sin. B. 



To prove (34). Let A o B (Fig. 9) -* A, and B o c = B ; then 

 A o c = A - B. 



In o c take any point P, and draw 

 the perpendiculars p Q, p B, and B 8, 

 B T, aa before (B T to p Q produced). 



_ QT PT 

 OP OP 



_QT_PT_B8_PT_BB OB 

 OP OP OP OP OB* OP 



_ PT P B 

 P B ' OP' 



Since the triangles T p B and o B s are equiangular, ' * = " 



Sin. (A - B) = ?* = 



PB 



.{../A T\_ BS OB 8 



sin. (A B) = . 

 OB O P OB 



PB 

 OP 



= sin. A eos. B - COB. A Bin. B. 



The above proofs evidently hold good only when neither of 

 the two angles exceeds a right angle. They can, however, be 

 extended to angles of any size by precisely similar construction, 

 which will, however, result in figures of very different appear- 

 ance, according to the quadrants in which the angles are situated. 

 In the demonstrations the minus sign belonging to sines and 

 cosines in certain quadrants \ a 



(see Section VIII.) must be 

 borne in mind. \ 



For instance, prove (33), 

 in the case where A and B 

 are both greater than right 

 angles, but where A -f- B is 

 less than three right angles. 

 Let AO B in Fig. 10 = A, and 

 B o c = B. In o c take any 

 point P as before, and con- 

 struct exactly as directed in 



j 

 / 



the proof for (33). 



Then, since A and B together form an angle in the third 

 quadrant, whose sine is a minus quantity 



Sin. (A + B) = - Z* = 

 OP 



OP 



B8 



OP 



_PT__ /R_8 OB\ _ /PJT PB\ 

 OP~ \^B OP/ VPB'OP/ 



Now A and B being both angles in the second quadrant, 

 their sines are both plus, and their cosines minus quantities. 



.'. = sin. A, for A o R = B o <J ; = - 

 OR o P 



o s 



= cos. A ; and 



p B 



sin. B; 



P T _ 



P B O R OP 



.'. sin. (A + B) = - (sin. A X - cos. B) - ( - cos. A X sin. B) 

 = sin. A cos. B + cos. A sin. B. 



Again, prove (34), where 

 A is a trigonometrical angle 

 in the fourth quadrant, B an 

 angle in the second quad- 

 rant, and their difference an 

 angle in the third quadrant. 



Let A o B in Fig. 11 = A, 

 and B o c = B ; /. A o c 

 = (A - B). Construct as 

 before 



Fig. 11. 



Sin. (A + B) = !_* 

 o P 



Q T , PT ^ B S , PT _ R8 

 OP OP OP OP~OB 



OP 



O B , PT PB 

 'OP P R ' O P* 

 PT 08 



Since the triangles T p Rand o RS are equiangular, = 



P B OR 



Thensin.(A-B)=- P<l = - P - T = 1-?- 



OP OP OP OP 



= R 8 _ P T _ B 8 OB _ PT F_R 

 OP OP OB'OP PB'OP* 

 But" 



8 = sin. A, for B o s = A o B ; -- - cos. B j 



OP 



O R 



O 8 



!_! = " B = cos. A ; 



PT 



P~R OR 



. . sin. (A B) = sin. A cos. B cos. A sin. B. 



and P B = sin. B; 

 o P 



