= 213=1+w3. It is always necessary that the value of one length at least should be given. A few examples are worked out below. c2=a+62 – 2ab cos C. Now cos C=cos 120°= - cos 60°= - ], ... C=7. a sin B 2a b= =as2=2, sin A 12 2 c=a cos B+b cos A = 1+. 2 Example 3. Given B=30°, b=3 J2, c=6, solve the triangle. csin B 6 1 .:. C=45° or 135o. 6 1/3 3/2 6/3 3/2 and a=ccos B+b cos C + 2 12 2 2 d1 Az are the two values of a, when B, b, c are given, show that ay + Az=2c cos B. We have ay=c cos B+b cos C1, and An=c cos B+b cos C2, where C1, C, are supplementary, so that cos C = -cos C2, adding a, +=2c cos B. = or .. EXAMPLES VII. = 1. Given A = 60°, b = 7 ft., C= = 5 yds., find a. 13, b = 3, c = 4, find A. c. /, = 7. = 8. = 9. Given a = 2, b = 12, c = 1/3 – 1, find A, B, C. sin 20° 36' 35" = 352, find A, B, C. 10. a 1 , 2 22 find c. = 11. Given sin C + cos C = find B. ✓ 17 find a :b : c. = 10 ft., b = 15 ft., C = 30°, find cot A and the area of ABC. 14. Given a=18, b=16, c= - 14, find the distance of B from the middle point of AC. 15. Given S = 6 sq. miles, a = 3 m., b = 5 m., Given (a + b + c)(b +c-a)= 3bc, find A. 18. If a cos A b cos B, show that the triangle is either isosceles or right-angled. 19. Given a = 193 ft., b = 194 ft., c=195 ft., find sin A, sin B, sin C. 20. If b cos A = = a cos B, show that the triangle is isosceles. 21. Given A = 75°, C = 45°, and the perpendicular from A on BC is 3 ft., solve the triangle. 22. Given tan C cot B= 5, tan B tan C = 9, and the distance of A from the middle point of BC is 7 in., find BC. If the base-angles of a triangle are 221° and 1121', show that its base is twice its height. 24. The sides of a triangle are in A. P., and the difference between the cotangents of the halves of the greatest and least angles is 4. Show that the cosine of the remaining angle is g. = 23. CHAPTER VIII. HEIGHTS AND DISTANCES. 175. The formulæ of Chapter VI. connecting the sides and angles of a triangle are of practical use in enabling us to calculate lengths and angles, which cannot be directly measured. Many objects in space are absolutely or practically inaccessible, though visible to the eye. In order to determine the distances of such objects from any others we require to measure directly (1) The distance between some pair of accessible points. (2) The angle between the lines drawn from the eye to some pair of visible points. By means of suitable instruments an angle can be measured with greater accuracy than a length. Hence it is usual to make our calculations depend on the measurement of a single length and of as many angles as are necessary. 176. The angle made with the horizontal plane by the line joining the eye to an object is called its angle of elevation [or, briefly, its elevation] when it is above the observer; its angle of depression (or, briefly, its depression) when it is below the observer. J. T. 8 177. To express the relation between the (vertical) height and the (horizontal) distance of an object. Let P be any object, A the position of the observer, PN the vertical line through P, and AN the horizontal line through A drawn to meet PN. (See first fig. of Art. 179.) The angle NAP is the angle of elevation of P, and the angle ANP is a right angle: thus AN cot NAP, .. distance = height (cot of elevation), NP NP tan NAP, .. height = distance < (tan of elevation). = These results are very important. Example 1. The angle of elevation of a tower 130 ft. high is observed to be 60°. Find the distance of the tower from the point of observation. Here the angle NAP=60°, and NP=130 ft., ... AN=NP.cot NAP=130 ft. x cot 60° 1 130 x 13 130 x 1.7320 =130 ft. x ft. ft. 3 Example 2. A man, standing on the top of a tower and looking in a direction perpendicular to the length of a river, which is 140 ft. wide, observes that the angles of depression of the two banks of the river are 45° and 30°. Find the height of the tower. Let A be the further, B the nearer bank of the river; P, the point of observation; and PN the vertical through P. (See fig. of Art. 179.) Then angle PBN=45°, and angle PAN=30°; and AB=140 ft. x + 140 ft. and i.e. =/3, +1 =cot 30°, ft. = 70 x 2:7320 ft. 13-1 3-1 =191•24 ft. = 1911 ft. nearly. .. = 178. The values of the trigonometrical ratios of any angles are to be found in published tables. If any ratio cannot be found by the elementary geometrical methods of Chapter IV. reference must be made to these tables. 179. To find the height and distance of an inaccessible object. P 4 N А B B Let P be the top of an object, PN the vertical through P. Let A, B be two points of observation, the distance between which is measured. (1) Suppose ABN to be in a straight line, so that A, B, N, P are in the same vertical plane. At A and B measure the angles of elevation NAP, NBP. At A measure the angles NAP, BAP; and at B measure the angle PBA. Then, both in (1) and (2), in the triangle PAB, the side AB and the angles PAB, PBA are known, .. AP can be calculated. .. in the right-angled triangle PAN, AP and įNAP are known, .. AN and PN can be calculated. |