point, and then there will be only one triangle answering the conditions. If the side CB is equal to the perpendicular Cd, the arc BB' will be tangent to ABB', and in this case also there will be but one triangle. When CB is less than the perpendicular Cd, the arc BB' will not intersect the base ABB', and in that case, no triangle can be formed, or it will be impossible to ful. fil the conditions of the problem. 2. Given two sides of a triangle 50 and 40 respectively, and the angle opposite the latter equal to 329 : required the remaining parts of the triangle. Ans. If the angle opposite the side 50 is acute, it is equal to 41° 28' 59"; the third angle is then equal to 106° 31' 01", and the third side to 72.368. If the angle opposite the side 50 is obtuse, it is equal to 138° 31' 01", the third angle to 9° 28' 59", and the remaining side to 12.436. CASE III. When the two sides and their included angle are given. 65. Let ABC be a triangle ; AB, B BC, the given sides, and B the given angle. Since B is known, we can find the sum of the two other angles : for A4 C ATC=180°-B and i(A+C)= (180°– B) We next find half the difference of the angles A and C by Theorem II. Viz. BC+BA: BC-BA:: tan }(A+C): tan }(A-C). in which we consider BC greater than BA, and therefore A is greater than C; since the greater angle must be opposite the greater side. Having found half the difference of A and C, by adding it to the half sum }(A+C), we obtain the greater angle, and by subtracting it from half the sum, we obtain the less. That is *(A+C)+ (A – C)=A, and Having found the angles A and C, the third side AC may be found by the proportion. sin A : sin B:: BC: AC. 1. In the triangle ABC, let BC=540, AB=450, and the included angle B=80°: required the remaining parts. INSTRUMENTALLY. Draw an indefinite right line BC and from any point, as B, lay off a distance BC=540. At B make the angle CBA=80°: draw BA and make the distance BA=450; draw AC; then will ABC be the required triangle. BY LOGARITHMS. = BC+BA=540+450=990; and BC-BA=540 – 450 =90. ATC=180°— B=180°—80°=100°, and therefore, i(A+C)= (1000)=50° . To find {(A,C). As BC+BA 990 ar. comp. 7.004365 : BC-BA 90 1.954243 :: tan }(A+C). 500 10.076187 : tan }(A-C). 6° 11' . 9.034795 Hence, 50° +6° 11'=56° 11'=A; and 50° — 6° 11'=43° 49'=C. . AB. : 640.082 2.806236 2. Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 04': required the other parts. Ans. Angles, 33° 34' 39" ; 18° 21' 21" ; side 2400. CASE IV. Having given the three sides of a plane triangle, to find the angles. greater side, dividing the given triangle into two right-angled triangles : then find the difference of the segments of the base by Theorem III. Half this difference being added to half the base, gives the greater segment; and, being subtracted from half the base, gives the less segment. Then, since, the greater segment belongs to the right-angled triangle having the greatest hypothenuse, we have the sides and right angle of two right-angled triangles, to find the acute angles. A 1. The sides of a plane triangle being given; viz. BC=40, AC =34 and AB=25: required the angles. B D INSTRUMENTALLY. With the three given lines as sides construct a triangle as in Problem IX. Then measure the angles of the triangle, either with the protractor or scale of chords. BY LOGARITHMS. As BC: AC +AB :: AC-AB: CD-BD In the triangle DAC, to find the angle DAC. ar. comp. 8.468521 26.6375 1.425493 :: sin D 900 10.000000 AC. . . BD. . . In the triangle BAD, to find the angle BAD. As AB 25. . ar, comp. 8.602060 : 13.3625. 1.125887 sin D. . 900 10.000000 : sin BAD 32° 18' 35" 9.727947 Hence 90° — DAC=90°-51° 34' 40"=38° 25' 20"=C and 90° - BAD=90° — 32° 18' 35" = 57° 41' 25"=B and BAD+DAC=51° 34' 40"+32° 18' 35"=83° 53' 15"=A. 2. In a triangle, in which the sides are 4, 5 and 6, what are the angles. ? Ans. 41° 24' 35" ; 55° 46' 16"; and 82° 49' 09". SOLUTION OF RIGHT-ANGLED TRIANGLES. 67. The unknown parts of a right-angled triangle may be found by either of the four last cases : or, if two of the sides are given, by means of the property that the square of the hypothenuse is equal to the sum of the squares of the other two sides. Or the parts may be found by Theorem V. EXAMPLES. 1. In a right-angled triangle BAC, there are given the hypothenuse BC =250, and the base AC=240: re- C quired the other parts. A To find the angle B. : . As BC 250 ar. comp. 7.602060 AC 240 2.380211 :: sin A. 900 10.000000 : sin B 73° 44' 23" 9.982271 But C=90-B=90° -73° 44' 23"=16° 15' 37" : |