We shall take the first of these cases, leaving the other as an exercise. Let then c2=(a+b)2(1-cos2 )=(a+b)2 sin2 0 ; .. c=(a+b) sin 0; .. log c=log (a+b)+ log sin 0. To determine we have the equation 1 2 .. log cos 0=log 2+ (log a+log b) − log (a+b)+log cos 2* Since 2√ab is never greater than a+b and cos is positive 2 and less than unity, cos is positive and less than unity, and thus is an acute angle. EXAMPLES. XVI. f. 1. If a=8, b=7, c=9, find the angles; given log 2, log 3, L tan 24° 5' 9.6502809, diff. for 60"-3390, L tan 36° 41′=9·8721123, diff. for 60′′=2637. 2. The difference between the angles at the base of a triangle is 24°, and the sides opposite these angles are 175 and 337 : find all the angles; given log 2, log 3, I tan 12° 9.3274745, Lcot 56° 6′ 27′′-9.8272293. 3. One of the sides of a right-angled triangle is two-sevenths of the hypotenuse: find the greater of the two acute angles; given log 2, log 7, L sin 14° 11'=9-455921, Lsin 14° 12′=9-456031. 4. Find the greatest side when two of the angles are 78° 14' and 71° 24′ and the sides joining them is 2183; given log 2·183=3390537, log 4.2274=6260733, D=103, 5. If b=2 ft. 6 in., c= =2 ft., A=22° 20', find the other angles; and then shew that the side a is very approximately 1 foot. Given log 2, log 3, L cot 11° 10' 10-70465, L sin 49° 27′ 34′′-9.88079, L sin 22° 20′ = 9.57977, L tan 29° 22′ 26′′=9.75041. 6. If a=1.56234, b=43766, C=58° 42′ 6′′, find A and B; given log 56234=4·75, given log cot 29° 21'=250015, log cot 29° 22′ = 249715. 7. If a=9, b=12, A=30°, find the values of c, having log 12-107918, L sin 30° =9:69897, log 9 95424, L sin 11° 48′ 39′′=9.31108, 8. The sides of a triangle are 9 and 3, and the difference of the angles opposite to them is 90°: find the angles; having given log 2, L tan 26° 33′ =9-6986847, L tan 26° 34′ = 9.6990006. 9. Two sides of a triangle are 1404 and 960 respectively, and an angle opposite to one of them is 32° 15': find the angle contained by the two sides; having given log 2, log 3, log 13=1.1139434, L cosec 32° 15′-10-2727724, L sin 21° 23' 9.5621316, 10. If b c 11 : 10 and A=35° 25′, use the formula L cos 24° 37′ 12′′=9.958607, L cot 17° 42′ 30′′= 10-495800, L tan 8° 28′ 56·5′′=9·173582, Ltan 12° 18′ 36′′ 9.338891, 11. If A=50°, b=1071, a=873, find B; given I sin 50° -9.884254, L sin 70° =9·972986, 12. If a=3, b=1, C=53° 7′ 48′′, find c without determining A and B; given log 2, log 25298=4.4030862, L cos 26° 33′ 54′′=9.9515452, (In the following Examples the necessary Logarithms must be taken from the Tables.) 13. Given a= = 1000, b=840, c=1258, find B. 14. Solve the triangle in which a=525, b=650, c=777. 15. Find the least angle when the sides are proportional to 4, 5, and 6. 16. If B=90°, AC=57·321, AB=28·58, find A and C. 17. Find the hypotenuse of a right-angled triangle in which the smallest angle is 18° 37′ 29′′ and the side opposite to it is 284 feet. 18. The sides of a triangle are 9 and 7 and the angle between them is 60°: find the other angles. 19. How long must a ladder be so that when inclined to the ground at an angle of 72° 15′ it may just reach a window 42.37 feet from the ground? 20. If a=31·95, b=21·96, C=35°, find A and B. 21. Find B, C, a when b=25·12, c=13·83, A = 47° 15′′. 22. Find the greatest angle of the triangle whose sides are 1837-2, 2385 6, 2173.84. 23. When a=21.352, b=45-6843, c=37·2134, find A, B, and C. 24. If b=647·324, c=850·273, A=103° 12′ 54′′, find the remaining parts. 25. If b=23-2783, A=37° 57′, B=43° 13′, find the remaining sides. 26. Find a and b when B=72° 43′ 25′′, C=47° 12′ 17′′, c=2484.3. 27. If AB=4517, AC=150, A=31° 30′′, find the remaining parts. 28. Find A, B, and b when a=324·68, c=421·73, C=35° 17′ 12′′. 29. Given a=321·7, c=435·6, À=36° 18′ 27′′, find C. 30. If b=1625, c= =1665, B=52° 19′, solve the obtuse-angled triangle to which the data belong. 31. If a=3795, B=73° 15′ 15′′, ('=42° 18′ 30′′, find the other sides. 32. Find the angles of the two triangles which have b=17, c=12, and C=43° 12′ 12′′. 33. Two sides of a triangle are 2.7402 ft. and 7401 ft. respectively, and contain an angle 59° 27′ 5′′: find the base and altitude of the triangle. 34. The difference between the angles at the base of a triangle is 17° 48′ and the sides subtending these angles are 105.25 ft. and 76-75 ft.: find the angle included by the given sides. 35. From the following data: (1) A=43° 15′, AB=36·5, BC=20, point out which solution is impossible and which ambiguous. Find the third side for the triangle the solution of which is neither impossible nor ambiguous. 36. In any triangle prove that c=(a - b) sec 6, where If a=17·32, b=13·47, C'=47° 13', find c without finding A and B. a+b C a-b 2' C 2 37. If tan = tan ;, prove that c=(a - b) cos seco. If a=27.3, b=16·8, C'=45° 12′, find 6, and thence find c. CHAPTER XVII. HEIGHTS AND DISTANCES. 198. Some easy cases of heights and distances depending only on the solution of right-angled triangles have been already dealt with in Chap. VI. The problems in the present chapter are of a more general character, and require for their solution some geometrical skill as well as a ready use of trigonometrical formulæ. Measurements in one plane. 199. To find the height and distance of an inaccessible object on a horizontal plane. Let A be the position of the observer, CP the object; from P draw PC perpendicular to the horizontal plane; then it is required to find PC and AC. α α B At A observe the angle of elevation PAC. Measure a base line AB in a direct line from A towards the object, and at B observe the angle of elevation PBC. Let LPAC=a, LPBC=ß, AB=a. PC-PB sin B. From A PBC, From A PAB, PC-a sin a sin ẞ cosec (ẞ − a). Also AC PC cot a=a cos a sin ẞ cosec (B-a). Each of the above expressions is adapted to logarithmic work; thus if PC=2, we have log x=loga + log sin a+ log sin ẞ+log cosec (B− a). |