B describe the Arc DC to cut the last blank line in the points D and C. Now if the angle C had been required obtuse, lines from D to B and to A, would constitute the triangle; but as it is required acute, draw the lines from C to B and to A, and the triangle ABC is constructed. From a line of chords let the angles B and C be measured, and AC from the same scale of equal parts that AB and BC were taken; and we have the answer required. Then 180-the sum of the angles A and C, gives the angle B Or 180°-(46° 30′ + 60° 31') = 72° 59′ = B. CASE 3rd. Having two sides, and the angle between them, to find the other two angles and the third side. RULE. As the sum of the two given sides, is to their difference, so is, the Tangent of half the sum of the unknown angles, to the Tangent of half their difference. Half the difference thus found added to half their sum, gives the greater of the two angles, and deducted, leaves the lesser. The third side is found by Case 1st. In the triangle ABC, there is given AB 240, the angle A 36° 40', and AC 180; to find the angles C and B, and the side BC. measure the angles C and B, and the side BC, as before; and we have the answer required. BY CALCULATION. The sum of the two given sides or ......... 240180 = 420 The difference 99 ......... 180° 36° 40′ or angle A = 240 180 = 60 143° 20' or sum of unknown angles B and C, the half sum of which is 71° 40′ The third side BC is found by Case 1st thus: = B As Sine of B: AC:: Sine of A: BC. If the angle included be a right angle, add the Radius to the Logarithm of the less side, and from the sum subtract the Logarithm of the greater side; the remainder or sum, will be the Tangent of the angle opposite to the less side. In the right-angled triangle ABC there is given AB 240, BC 180, and the angle B 90° to find the angles A and C and the side AC. Then, To the Logarithm of less side or 180 And deduct the Logarithm of greater side or 240... 2.3802112 Tangent of A= 36° 52'......... answering to... Then 180° (90° 00'+36° 52') = 53° 08' And, As Sine of C: AB:: Sine of B : AC. CASE 4th: Having the three sides, to find any angle. RULE. As the longest side or base, is to the sum of the other two sides, so is the difference of those sides, to the difference of the segments of the base, formed by a perpendicular let fall from its opposite angle. Half the difference of the segments thus found, added to half the base gives the greater segment, and subtracted leaves the lesser-whence we have two rightangled triangles with two sides given, to find the other two angles and the third side, by Cases 1st and 2nd. In the triangle ABC, there is given AB 64, AC 47, BC 34: to find the angles A, B, and C. GEOMETRICALLY. Draw a blank line, on which from a scale of equal parts lay off AB 64, from A with the line AC 47, describe an arc, and from B with the line BC 34 describe another arc, to the A 64 intersection point at C, draw AC and BC and ABC is the triangle required. BY CALCULATION. From the point C, let fall the perpendicular CD, on the base AB; it will divide the triangle into two right-angled ones ADC and CBD, as well as the base AB into the two segments AD and DB, The longest side or base AB = 64. The sum of the other two sides or AC + BC = 47 + 34 = 81 ,,AC — BC : AD-DB 16.46......... answering to وو The half difference or 8.23 added to the half base or 32 = 40.23 or greater segment, and deducted = 23.77 or lesser segment. Having now two right-angled triangles ADC and CBD with two sides given in each, the remaining angles and side are easily found by Cases 1st and 2nd. HEIGHTS AND DISTANCES. Having given the solutions of the four cases in Plane Trigonometry, we will proceed to apply the same, by a few examples, in determining the heights and distances of inaccessible objects. EXAMPLE 1. What is the height of a Musjid, the distance of which was 384.40 yards from where I stood, and the angle of elevation to the top 3° 42'? |