That is to say, the sum of two of the sides of a plane triangle is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. 76 This proportion is employed when two sides and the included angle of a triangle are given to find the other parts. Since the three angles of every triangle are together equal to two right angles or 1800, subtracting the given included angle from 180°, the remainder is the sum of the two angles opposite the given sides; then substituting for a and b in the above. proportion the two given sides, three terms of it are known and the fourth may be found. After which, having half the sum and half the difference of the unknown angles, these angles themselves can be found by adding half the sum to half the difference for the greater, and subtracting half the difference from half the sum from the lesser; when all the parts of the triangle will be known, except one side, which may be found by the proportion the sines of the angles are as the opposite sides. (Art. 67.) EXAMPLE. An observer wishing to know the length of a small lake, measured two lines from the same point to the two extremities of the lake, which he found to be respectively 153 and 137 yards; he also observed with an instrument for taking angles the angle subtended from this point by the lake to be 40° 33' 12". 153yds 4033′12' 137 yda The blank form for this case would be as follows: * This notation "log.-" signifies the number whose log. is. † A more direct mode of finding this side when it is the only part required, is given at Art. 77. 153 is known to be the side opposite 78° 13' 1'' because the greater angle of a triangle is always opposite the greater side. (Geom. Theorem 9.) The same form filled up with the given example above is given below. 77. Given two sides and the included angle of a plane triangle to determine the third side, without finding the remaining angles. The general expression for the side c, in terms of the two sides a, b, and the included angle c, is (Art. 69) on the supposition of R = 1, Assume the second term within the brackets equal to tan 20, then, since 1+ tan ze Hence c is determined by these two formulas, viz., log. tan 0 = log. 2+ log. a+ log. b + log. sinc-log (a —b) * Log. cot } c might be used instead of log. tan§ (A + B) since 180° A+B and ... 90° — § c = † (A + B) and tan (90° — § c) =cotc. But nothing would be gained, since (A+B) must be employed to add and subtract with (A — B) + When b is nearly equal to a, the following formulas will give c with greater exactness. For demonstration see App. I. Art. 23. Given a 5891, b 4562·34, c 30° 20' 10"-3 to find the other parts of the triangle. Ans. A 99° 57' 5''•25, в 49° 42′ 44.45, c 3020.823. Given a 40° 55′ 31', b 83-25, c 100, to find a. Ans. a 65.9574. This case of the solution of a triangle combined with that exhibited at Art. 67, serves to determine the horizontal distance between two inaccessi ble objects. Let the distance between two towns which are in sight be required. Measure a line upon the ground (which is called the base line) of 2 miles. Take the angles at each extremity formed by this base line and a line to each of the towns. Two triangles will be formed, in each of which a side, viz. the base line, and two adjacent angles will be given. Let the angles in the triangle of which the upper town is the vertex be 159° and 14°; and those in that of which the lower town is the vertex be 25° and 149°. Calculate the distance from one extremity of the base line, say the upper extremity, to each of the towns, as in Art. 67. Then 2 Miles 1490 you will know two sides of a triangle the third side of which is the dis tance between the towns required. The included angle between these two sides is 159° 25° 134°. Having then two sides and the included angle, the remainder of the solution is the same as in the last case. We leave it as an exercise for the learner. Ans. 13.08 miles. MISCELLANEOUS EXAMPLES. (1.) From the equation sin2 a +5 cos2 a = 3 to find the value of sin a. (2.) If sin2 a = m cos a―n to determine cos a. Ans, sin a= √2 Ans. cos a =— - } m ± √ ‡m2 + n + 1. (3.) Given sin a = m sin b, tan a = n tan b to find sin a and cos b. (7.) Given the base of a triangle 87.75, the vertical angle 73° 20′, and the difference of the angles at the base 13° 4′, to find the other parts. (8.) Given the base 117.3, the vertical angle 19° 18', and the ratio of the other two sides 8: 11 to solve the triangle. Aus. 123° 13' 23.7, 37° 28′ 36'3, 215.94, 296.89. (9.) Find into what two parts the perpendicular from the vertex of the angle c, divides the side c of a triangle when a = = 33°, c = 75°, and a = 2134. Ans. 3125-2 and 659.44. (10.) The side of a regular polygon of 41 sides is 0-736. What is the ratio of the radii of the inscribed and circumscribed circles? Ans. 99708. (11.) Find the extent of the circle of vision, viz. the arc To, from the top of a mountain м, whose height is 5460 feet, supposing the radius of the earth to be 3956.1 miles. Ans. 91 miles, 1558.88 feet. T M (12.) When the sun's altitude is 19° 16', the peak of a mountain casts its shadow at a certain point, and at another point when the sun's altitude is 20° 42', distant from the former point 937 ft. Now supposing the sun to be vertical on the top of the mountain at noon, what is its height? Ans. 4369.1 ft. |