then if AB be the required distance, we shall obtain it by the following proportion. sin B sin C:: AC: A B. : Example III. To find the distance between two inaccessible objects, B and C (fig. 13.), we must measure a base as A D, and observe the angles at each extremity which both objects make with this line. Suppose we have found A D=743 · 55 yards, BAD=113° 44′, C A D=48o, A D C=104° 13′, and B D A 37° 8'; we shall then get the angles ABD 29° 8′ and ACD=27° 47'. Hence we shall have A B and A C by these proportions, sin A B D : sin A DB:: AD: AB sin A CD: sin ADC :: AD: A C performing the computations, A B=921980 and A C=1546 30. Now, in the triangle A B C, the two sides A B, AC are known, with the included angle BAC-DAB-D A C*-65° 44', to find • We have supposed the four points A, B, C, D, to lie in the same plane; when this condition does not hold, the angle BAC will no longer be the difference between D A B and DAC: in this case it will be necessary to have the value of that angle by direct measurement; in other respects the process will be the same. Now to find BC we have, şin A CB: sin B A C :: A B : B C. log sin 35° 45′ 14′′ ..... 9.7666394 log B C... 31579069 Hence the required distance B C=14385 yards. Example IV. Three points A, B, C (fig. 14.) in the map of a country being given, it is required to determine the position of a fourth point S, and the three distances AS, BS, CS; the angle AS B, BSC being known, and the four points lying in the same plane. Describe a circle about the triangle A C S, which will intersect BS in D; join A D, D C. Since angles in the same segment of a circle are equal, DAC DS C and A C D AS D. Hence in the triangle A D C, the angles A and C are known and the side A C, therefore A D may be found. In the triangle ABC the three sides are known, hence the angle B A C may be determined, and thence B A D. Now, in the triangle A B D, the two sides A B, AD E are given with the included angle, to find the angle ABD: then the angles A S B, A B S, and the side A B being known, the distances AS, BS may be found. Lastly, in the triangle A C S all the angles and the side A C are given, to find CS. The problem may be solved geometrically by describing upon A B a segment of a circle containing an angle equal to A S B, and upon BC a segment containing one equal to BSC. The intersection of the arcs will determine the position of the point S. (62.) To find the area of a triangle which has the three sides given. Of the triangle A B C (fig. 8.), Hence, = S. (S-a). (S-b). (S—c). log area logAC+log A B+log sin A-10. or (log S+log S-a+log S-b+log S―c). (63.) To find the radius of the circle inscribed in a given triangle. Let A B C be the triangle (fig. 15.); bisect the angles B A C, A C B by the straight lines A O, C O; the point of intersection O, will be the centre of the inscribed circle (Euc. IV. 4). Draw OR, OP, O Q, perpendicular to the sides of the triangle; join O B, and suppose r to denote the required radius: then, a Area of ▲ BOC=: ВС 2 By addition, we have, area of ▲ ABC=a+b+c r = 8r, 2 or, √S. (S—a) · (S—b) . (S—c)=Sr, (S—a) · (S—b) · (S—c) S (64.) To find the radius of the circumscribed circle. Suppose A B C (fig. 16.) to be the triangle; bi |