VOL. XV.] PHILOSOPHICAL TRA^SACTIONS. 23/ 



i 'If the stations A, E fall in a right Hne with the point C; the lines G A, HE, 

 being parallel, cannot meet : but in this case the problenn is indeterminate, and 

 capable of infinite solutions. For, as before, on C G describe a segment of a 

 circle capable of the observed angle cea; and upon CH, describe a segment 

 capable of the observed cae : then through C, draw a line any way cutting the 

 circles in A, E, these points will answer the question. 



Prob. 3. — Four points, BCDF, (fig. 15), or the 4 sides of a quadrilateral, 

 with the angles comprehended, are given ; also there are 2 stations A and E 

 such, that at A, only B, C, E are visible, and at E only A, D, F, that is, the 

 angles BAG, B A E, A E D, D E F are given : to find the places of the two 

 points A, E; and consequently, the lengths of the lines AB, AC, AE, 

 ED, EF. 



Construction. — Upon BC describe a segment of a circle, that may contain an 

 angle equal to the observed angle BAG; then from G draw the chord C M, or 

 a line cutting the circle in M, so that the angle B C M may be equal to the 

 supplement of the observed angle B A E, i.e. its residue to 180 degrees. In 

 like manner, on D F describe a segment of a circle, capable of an angle equal 

 to the observed D E F ; and from D draw the chord D N, so that the angle 

 F D N may be equal to the supplement of the observed angle A E F ; join M N, 

 cutting the 2 circles in A, E : then A, E, are the two points required. 



Demonstratioti. — Join A B, A G, E D, E F ; then is the angle M A B = B C M 

 (by 21, 3, Eucl.) = supplement of the observed B AE by construction ; there- 

 fore the constructed angle B A E is equal to that which was observed. Also the 

 angle B A G of the segment is, by construction of the segment, equal to the 

 observed angle BAG. In like manner the constructed angles, A E F and D E F, 

 are equal to the correspondent observed angles, A E F, D E F ; therefore A, E 

 are the points required. 



Calculation. -^\n the triangle B G M, the Z. B G M (= supplement of B A E) 

 andZ.BMG (=BAG) are given, with the side BG; thence MGmaybe 

 found ; in like manner D N, in the A D N F, may be found. But the Z. M G D 

 (= B G D — B G M) is known, with its legs M G, G D, therefore its base M D 

 and Z. M D G may be known. Therefore theZ.MDN(=GDF — GDM — 

 F D N) is known, with its legs M D, D N ; thence M N, with the angles D M N, 

 D N M will be known. Then the Z. G M A (= Z. D M G + D M N) is known, 

 with the Z. M A G (= M A B + B A G) and M G before found ; therefore M A 

 and A G will be known. In like manner, in the triangle E D N, the angles 

 E, N, with the side D N, being known, the sides E N, E D will be known ; 

 therefore AE(=MN — MA— EN)is known. Also in the triangle A B G, 

 the Z. A, with its sides B G, C A being known, the side A B will be known. 



