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withms. For, since the given angle is 90°, the half sum of the remaining angles is 45°, whose natural tangent is unity; and the sum and the difference of the given sides are 72 and 8 respectively. Hence - 8 l As 72:8; 1 ; ; = , = 1111111 = nat. tan. 6°20'. Therefore 45°-H 6°20′ = 51° 20', and 45° – 6°20' = 38° 40', are the remaining angles. And the third side (Euc. i. 47) = M403 ± 32= V2324 = 8 v4.1 = 51.225. *** Other methods, still shorter, of solving right angled triangles, will be given before we terminate the present chapter. Ev. 3. Given two sides of a plane triangle 1686, and 960, and their included angle 128° 4'; to find the rest, Ains. Angles 33° 35', 18° 21', side 2400,
10. When the three sides of a plane triangle are given, to find the angles. 1st Method. Assume the longest of the three sides as base, then say, conformably with chap. ii. prop. 16. As the base, To the sum of the two other sides; So is the difference of those sides, To the difference of the segments of the base. Half the base added to the said difference, gives the greater segment, and made less by it gives the less; and thus, by means of the perpendicular from the vertical angle, divides the original triangle into two, each of which falls under the first case. 2d Method. Find any one of the angles by means of prop. 17, of the preceding chapter; and the remaining angles either by a repetition of the same rule, or by the relation of sides to the sines of their opposite angles.
RIGHT ANGLED PLANE TRIANGLES.
12. 1. Right angled triangles may, as well as others, be solved by means of the rule to the respective case under which any specified example falls: and it will then be found, since a right angle is ...; one of the data, that the rule usually becomes simplified in its application; as appeared in the solution of the second example to case 2. 2. When two of the sides are given, the third may be found by means of the property demonstrated in Euc. i. 47. Thus, Hypoth. = y (base" + perp.”.) Base = V (hyp.”— perp.”) = W(hyp. + perp.). (hyp. – perp.) Perp. = V (hyp.” - base") = x/úly. Ibass). UNT. E. base). 8. There is another method for right angled triangles, known by the phrase making any side radius; which is this. “To find a side, Call any one of the sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write those words upon them accordingly. Call the word written upon each side the name of each side: then say, “As the name of the given side, “Is to the given side; “So is the name of the required side, “To the required side.” “To find an angle. Call either of the given sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write those words on them accordingly. Call the word written upon each side the name of that side. Then say, “As the side made radius, “Is to radius; “So is the other given side, “To the name of that side: which determines the opposite angle.”
13. When the numbers which measure the sides of the triangle, are either under 12, or resolvable into factors which are each less than 12, the solution may be obtained, conformably with this rule, easier without logarithms than with them. For,
Let ABC be a right angled triangle, in C which AB the base is assumed to be radius; t Bc is the tangent of A, and Ac its secant, to that radius; or, dividing each of these by the base, we shall have the tangent and A. . . B
secant of A, respectively, to radius 1. Tracing in like manner, the consequences of assuming Bc, and Ac, each for radius, we shall readily obtain these expressions.
1. # = tan angle at base.
hyp. 25 l()0 Ex. 2. In a right angled triangle, given a leg and its opposite angle 384, and 53° 8' respectively; to find the other leg and the hypothenuse. Ans. Leg 280, hypothenuse 480. Ev. 3. In a right angled triangle are given the base 195, its adjacent angle 47° 55'; to find the rest. Ans. Perp. 216, hyp. 291, vert. ang. 42° 5'.
Plane Trigonometry considered Analytically.
1. IN the preceding chapters the investigation of trigonometrical properties has been conducted geometrically; the various relations of the sines, cosines, tangents, &c. of arcs and angles, whether depending upon triangles or not, being deduced immediately from the figures to which the several enquiries were referred. This method carries conviction at every step; and by keeping the objects of enquiry constantly before the