of the Compass it is to be set off in as also whether it be nearer the East or West than the North or South for if it be nearer the East or West than the North or South you must do the same from a Parallel as here you did from a Meridian for the side of a Square is but the measure of four points which is but half the points between a Meridian and a Parallel This way of working may seem hard and tedious at first but you will soon find that it is free from mistakes and both exact and easie if you practise it Place this between Page 112 and 113. Of OBLIQUE TRIANGLES Two Sides with an Angle opposite to one of them given to find the other Angles and Side QUESTION I. Two Ships set sail from the Rock of Lisbon one sailed W S W the other sailed N W b W 38 leagues and at the end of their sailing they were 58 leagues asunder I demand the Southermost Ships Distance run and how the Ships bear one from the other FIrst draw a North and South line white and then from that set off the Northermost Ships Course make the Rock or Place setting out the Place in the North and South line that you draw that Course from which is C upon this Course set off 38 leagues because the Question saith you sailed 38 leagues upon it and extend the side C A to the Arch of 60 deg at t and set off five points from t to S and draw S C a white line which is W S W Course for it is five points between the W S W and the N W b W this done take 58 leagues from the Scale of equal parts and fix one foot of your Compasses in A and where the other intersects the Course B C which it doth in B there is the other Ship then to measure the Angle of the Ships bearing one from the other it is B and B C is an E N E line extend B A to the Arch of 60 deg whose Center is at B and see how many degrees or points it is more Northerly than an E N E line and so the other Ship namely the Ship at A bears from the Ship at B then take the length of B C and apply it to your Scale and see how many leagues or miles it is I have wrought this in leagues but I will work the rest in miles because it is more exact I find that the Ship at A bears from B N E b N 45 min. Easterly the Distance run of the Southermost Ship is 70 leagues By the Tables The proportion of this and all others of this kind is the same that holds in right angled Triangles namely that the Sine of every Angle is proportional to its opposite side or every side is proportional to the Sine of its opposite Angle Here we have given the side C A 38 Leagues the side A B 58 Leagues and the Angle at C which is 5 points or 56 deg 15 min. Say then for the Angle at B. As A B 58 leagues 580 tem comp arith 7,2365719 Is to A C B 56 deg 15 min. Sine  9,9198464 So is A C 38 leagues 380 tem  2,5797836 To A B C Sine 33 deg 0 min.  9,7362019 Which subtract from 6 points or 67 d. 30 m.  33 00 Remainder is the Course from North 34 d. 30 m. Which the Ship A bears from the Ship B which is N E b N 45 min. Easterly For the Distance run of the Ship at B the Side B C. As Sine C 56 deg 15 min. comp arith 0,080153 Is to A B 58 leagues  2,763428 So is Sine B A C 90 deg 45 min. Take the Sine of the acute Angle B A t 89 d. 15 m. 9,999961 To B C 68 8 10  2,843542 The reason why you take the acute Angle is because the Tables go no further than 90 deg neither indeed is any Sine beyond 90 deg but as my Father saith in his Trigonometry p. 2 the Sine of an Arch less than a Quadrant is also the right Sine of an Arch as much greater than a Quadrant So then the right Sine of the Arch 90 deg 45 min. which is 45 min. greater than a Quadrant must be the right Sine of the Arch 89 deg 15 min. which is as much less namely 45 min. the Geometrical Demonstration of it is there laid down You may ask how I came to know the Angle B A C which was thus I found one of the other two Angles namely B and the other I had given me I added them both together and that Sum I subtracted from 180 deg the Remainder must then be the Angle at A because 180 deg is the Sum of the three Angles of any right lined Triangle And if I subtract two of them from three of them there will remain one of them which was 90 deg 45 min. The three Angles of a Triangle given with one of the Sides to find the other two sides QUESTION II. Admit I set from a head-Head-land lying in the Latitude of 50 deg 00 min. North Latitude and sail W S W 38 Leagues and then meet with a Ship that came from a Place which lies S S W from the head-Head-land Now this Ship hath Sailed N W. I demand the Distance of that Place from the other and also the Distance the Ship hath sailed that came from the Southermost Place The distance between the two places is A B 58 miles The distance that the Ship sailed is A O 44 4 10 miles If you have occasion to find the Latitude the place at A lies in let fall the Perpendicular A N upon the South line that comes from B and it cuts it in N and leaves the Difference of Latitude B N see how many miles it is and subtract it from 50 deg and you have your desire measure A N and you have the Westing that A lies from the Head-land B. By the Tables Here the Angle at B is an Angle of 4 points or 45 d. 00 m. The Angle at A is an Angle of 6 points or 67 d. 30 m. The Angle at O is an Angle of 6 points or 67 d. 30 m. The side B O is 58 miles   First for the Side A B. The general Rule saith That the Sine of every Angle is proportional to its opposite side Then from this I conclude that B A should be equal to B O because the Angles opposite to them are equal and so you will find them For as Sine A 67 deg 30 min. comp arith 0,034384 Is to O B 58 miles  1,763428 So is Sine O 67 deg 30 min.  9,965615 To A B 58 miles the two places distance 1,763428 You see it is so exactly for these three numbers added together and Radius cast away produceth the same Logarithm that 58 taken out of the Book did and this Question I do on