5. Given B = 71° 41' 10", c = 24, find b. = = Log 24 1.3802112, L cos 18° 18' 9.9774609, = 12. CD is the perpendicular from C on AB, and DE the perpendicular from D on BC. Given B = 60°, a = find DE. 20, CHAPTER IX. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three sides of a triangle, to find the remaining may be determined; and from similar formulæ we may find B and C. These formulæ are not however adapted to logarithmic computation. We shall therefore find it generally advisable to use the formulæ of Art. 34. From either of these formulæ we can determine A, and from similar formulæ determine the other angles. 39. Given one side and two angles, to find the remaining parts. Of course the third angle is at once known. Let a be the given side. both of which formula are adapted to logarithmic computation. 40. Given two sides and the included angle, to find the remaining parts. Let b, c be the given sides, and A the included angle. NOTE.-When the two given sides are equal, the solution may be effected more easily by drawing a perpendicular from the given angle upon the opposite side, and so bisecting it. By drawing a figure, it is easily seen that, in this case, B = C = 90° – A, and a 2b cos B. 41. Given two sides and an angle opposite to one of them, to find the remaining parts. Let a, b, B be the given elements. We then have sin A> 1, which is impossible. Hence, in this case, it is impossible to form a triangle with the given elements. Then, since, Art. 15, the sine of an angle is the same as the sine of its supplement, there are two values of A which α satisfy the equality, sin A = sin B, and these values are supplementary. Let A, A' be the two values, then the relation between them is A+ A'· = 180°. If a is not greater than b, then A is not greater than B, and there is no doubt as to which value of A is to be taken. If, however, a is greater than b-that is, if the given angle is opposite to the less of the given sides, we must have A greater than B, and both values of A may satisfy this condition. This particular case, when the given angle is opposite to the less of the given sides, is called the ambiguous case. We will illustrate this geometrically. 42. The ambiguous case. Let a, b, B be given to construct the triangle. given side a, and draw BA making Then with centre C and radius = = B. = a, AC = b, and == a, Again, the sides BA and BA' correspond to the two values of c which are obtained from the two values of A in the And the angles ACB and A'CB correspond to the two values of C which would also be found. COR. If a perpendicular CD be drawn from C upon AA', and if c' and c be the lengths of BA' and BA respectively, it may be easily shown that c' + c 2 a cos B, and c' c = 2 b cos A. = 26° 32', and ab:: 3:5, find A, B. Log 2 3010300, L cot 13° 16' = = 10.6275008, L tan 46° 40' = 10.0252805, L tan 46° 41' = 10·0255336. = = and c. 11. a = 3, b 2 A 60°, find B, C, Log 23010300, log 3 4771213, = = L sin 35° 15' 9-7612851, L sin 35° 16' 9.7614638, 12. a = Log 2 Log 3 = =3 = = 6, с 7, find A. 13. If c, c' be the two values of the third side in the ambiguous case when a, b, A are given, show that 14. If a, b, A are given, show from the equationb2 + c2 2 bc cos A = a2; that if c and c' be the two values of the third side 15. Show also from the same equation that there is no ambiguous case when a = b sin A, and that c is impossible when a < b sin A. 16. Having a b, A, B, solve the triangle. 17. Given the ratios of the sides, and the angle A, solve the triangle. A 18. If in a triangle tan § (B-C) = tan cot, show that b cos p = C. CHAPTER X. HEIGHTS AND DISTANCES. 43. We shall now show how the principles of the previous chapters are practically applied in determining heights and distances. We have not space here to describe the instruments by which angles are practically measured, but we shall assume that they can be measured to almost any degree of accuracy. |