GEOMETRICAL CONSTRUCTION.

From a scale of equal parts draw A B = 276.5, and by the protractor, or line of chords, make angle A 59°. Now the triangle cannot be constructed without having the angle B j therefore subtract the sum of angles A and c from 180° which gives angle n, that is 1S'0—(59°+52° 15') or 180-111° 15' 68.45 = angle B ; make angle B 68° 45', then A c and 13 c being measured by the same scale as A B, will give their re spective lengths.

Case sides, and an angle opposite to one of them, being given ; to find the other two angles and the remaining side.

As the side opposite the given angle is to the side opposite the required angle so is the given angle to the required angle.

the oblique-angled triangle A B c, obtuse at B, given A c 318 yards, B c 195 yards, and the angle A 32° 40', to find the angles B and c, and the side A B.

A c = 31S from a scale of equal parts ; make angle A = 32° 40' ; with the distance 195 equal parts and the centre c describe an arc cutting A Bat a, and join B c; the angles B and c will be found by the protractor, or line of chords, and the side A B by the same scale from which the other two sides were taken.

By Calculation. As the side B c, 195 2.290035 is to the side A c, 318. 2.502427 so is the sine' A, 32° 40' 9.732193 12.234620 to the sine of a, 61° 40' 9.944585 But since the tables give only acute angles, and the angle required is obtuse, and the sine of any angle is the same as the sine of its supplement, therefore 180-61° 40' = 11S° 20' = angle B.

the given side opposite to the given angle is greater than the other given side, the angle opposite to such other given side, or the angle to be first found, is always acute, and is found by proportion ; but when the side oppo site to the given angle is less than the other given side, the opposite angle may either be acute or obtuse.

Case the two sides and the included angle ; to find the other two angles, and the third side.

As the sum of the two sides is to their difference, so is the tangent of half the sum of the opposite angles to the tangent of half their difference.

Having by this proportion found the difference of the angles of the base, then half of the sum added to the half difference, gives the greater angle, and half of the sum dimin ished by the half difference gives the lesser angle.

Example.—For the triangle ABC, given the side A c 919.95, the side A B 500 feet, and the contained angle A 52'; to find the angles B and c, and the side B C.

the angle B A c = 52' ; make A B = 500 from a scale of equal parts, which set from A to B; from the same scale transfer 919.95 from A to c, and join A C ; then the side A c will be found upon the line of equal parts, and the angles a and c by the protractor, or line of chords.

12.477089 to B c, 600.26 2.77833S Case Given the three sides of a triangle, to find the angles.

From the angular point opposite the greater side, draw a perpendicular to that side, dividing it into two segments ; then As the base, or sum of the two segments of the base, is to the sum of the other two sides, so is the difference of the sides to that of the segments of the base : add half the sum of the segments of the base, and half their difference will give the greater segment : while the half sum subtracted from half the difference, will give the lesser seg ment.

Example.—In the triangle A B C, given the side A B, 562, A C 800, and a C 320 ; to find the angles.

the straight line A c = 800 equal parts from any scale ; with a radius of 562 equal parts, and the centre A describe an arc at B, and with a radius of 320 and the centre c' describe another are, cutting the former at B; join A B and B c; then measure the angles by a protrac tor, or line of chords.

Now AB +BC =562 + 320 = SS2 the sum of the sides, and A B — B c=562-320=242 their difference ; then As the base, or longest side, 800 903090 gives the whole angle AB c = 4' then 90— 39'=1S° 21' = angle A and 90 — 56° 25'=33° 35' = angle c.

From three parts of a triangle, including at least one side as a part given, the other three remaining parts have been found upon general principles; but when the triangle is right angled, other solutions, more simple for general practice, may be given ; these are founded upon the following proportions : Let A B C be a right-angled triangle, right-angled at B ; upon A B take any radius, A D ; with A as a centre, and the radius A B, describe an arc, D o, cutting A c at o ; draw D E and o F parallel ton c, cutting A c at E, and A B at F ; then, by the definitions given, A c or A D is the radius, E D is the tangent, o F the sine, A F the cosine, and A E the secant of the arc c D. Now the sine, cosine, tangent, &c. of an arc, being the same as the sine, cosine, tangent, &c. of an angle drawn from the centre through the extremities of the arc, the following proportions result by similar triangles, A D E and A B C.