It must be remembered, however, that in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A. In the same manner the cases in which A is equal to a right angle or greater than a right angle may be discussed, and the following results obtained. As before in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A. It will be seen from the above investigations that if a lies between b and π-b, there will be one solution; if a does not lie between b and b either there are two solutions or there is no solution; this enunciation is not meant to include the cases in which a=b or π - b. 87. The results of the preceding article may be illustrated by Let ADA'E be a great circle; suppose PA and PA' the projections on the plane of this circle of arcs which are each equal to b and inclined at an angle A to ADA'; let PD and PE be the projections of the least and greatest distances of P from the great circle (see Art. 59). Thus the figure supposes A and b each less than π 2' If a be less than the arc which is represented by PD there is no triangle; if a be between PD and PA in magnitude, there are two triangles, since B will fall on ADA', and we have two triangles BPA and BPA'; if a be between PA and PH there will be only one triangle, as B will fall on A'H or AH', and the triangle will be either APB with B between A' and H, or else A'PB with B between A and H'; but these two triangles are symmetrically equal (Art. 57); if a be greater than PH there will be no triangle. The figure will easily serve for all the cases; thus if A is greater we can suppose PAE and PA'E to be equal to A; if we can take PH and PH' to represent b. 2' 88. The ambiguities which occur in the last case in the solution of oblique-angled triangles (Art. 85) may be discussed in the same manner as those in Art. 86; or by means of the polar triangle, the last case may be deduced from that of Art. 86. EXAMPLES. 1. The sides of a triangle are 105°, 90°, and 75° respectively; find the sines of all the angles. 2. Shew that tan A tan B Sin (8-c) Solve a triangle = sin s when a side, an adjacent angle, and the sum of the other two sides are given. 3. Solve a triangle having given a side, an adjacent angle, and the sum of the other two angles. 4. A triangle has the sum of two sides equal to a semicircumference; find the arc joining the vertex with the middle of the base. 5. If a, b, c are known, c being a quadrant, determine the angles; shew also that if & be the perpendicular on c from the opposite angle, cos2d = cos2 a + cos2 b. 6. If one side of a spherical triangle be divided into four equal parts, and 0,, 0, 0, 0, be the angles subtended at the opposite angle by the parts taken in order, shew that 4 sin (0, + 0,) sin 0, sin 0 = sin (0 ̧+ 01) sin 0, sin 0 ̧. 2 3 7. In a spherical triangle if A= B= 2C, shew that 8 sin (a+) sin' cos-sin" a. = 8. In a spherical triangle if A = B = 2C, shew that 9. If the equal sides of an isosceles triangle ABC be bisected by an arc DE, and BC be the base, shew that 10. If c,, c, be the two values of the third side when A, a, b are given and the triangle is ambiguous, shew that VII. CIRCUMSCRIBED AND INSCRIBED CIRCLES. 89. To find the angular radius of the small circle inscribed in a given triangle. Let ABC be the triangle; bisect the angles A and B by arcs meeting at P; from P draw PD, PE, PF perpendicular to the sides. Then it may be shewn that PD, PE, PF are all equal; Hence BC+ AF = half also that AE AF, BF= BD, CD=CE. = the sum of the sides Now = s; therefore AF = 8 α. Let PF= tan PF = tan PAF sin AF (Art. 62); The value of tanr may be expressed in various forms; thus from Art. 45, we obtain tan r = √{cos S cos (S- A) cos (S – B) cos (S′ – C)} 2 cos A cos B cos C 4 cos 4 cos B cos C = cos S + cos (S− A) + cos (S′ – B) + cos(S–C); hence we have from (4) + cos (S – A) + cos (S – B) + cos (S - ......... .(5). |