Notebook: Roger B. Farquhar, 1852-1853

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Practical Questions

Example 1st At 85 feet distance from the bottom of a lower the angle of its elavation was found to be 52° 30' required the altitude of the lower. Ans 110.8 ft

[Diagram of a right triangle, with angle ABC being the right angle]

As sine of L C 37.30 ---- 0.21555 As to sine of L A 52.39 ---- 9.89947 So is given side AB 85 ---- 1.92942 To required side CB 110.8 ---- 2.04444

Example 2nd

To find the distance of an inaccessible object I measured a line of >3 yards and at each end of it took the angle of position of the object and the other end and found the one to be 90° and the other 61° 45' required the distance of the object from each station. Ans 135.9 years from one and 154.2 from the other.

[diagram of a right triangle, with angle ABC being the right angle]

As sine of L C 28°.15 ---- 0.32485 1st to sine of L B 90° ---- 10.00000 So is given side AB > 3. ---- 1.86332 To required side AC 154.2 ---- 2.18817

As sine of L C 28.15 ---- 0.32485 Is to sine of L A 61.45 ---- 5.54452 So is given side AB >3. ---- 1.86332 To required side BC 135.9 ---- 1.13309

Example 3rd Wishing to know the distance between two trees C and D standing in a fog I measured a base line AB 339 feet at A the angle BAD was 100° and BAC 36° 30' at B the angle ABC was 121° and ABD 49° required the distance between the trees Ans 697 1/2 feet

[diagram of a trapezoid, with line DC and AB forming the two parallel bases - line DC is longer than line AB, and two lines (DB and CA) are drawn connect opposite ends of the trapezoid]

First work in triangle ABD to find side AD As sine L D 31 ---- 0.28816 Is to sine L B 48 ---- 9.87778 So is given side AB 335 ---- 2.53020 To required side AD 496.7 ---- 2.69614

Second work in triangle ABC to find side AC As sine L C 22° 30' ---- 0.41716 Is to sine L B 121° ---- 9.93307 So is given side AB 339 ---- 2.53020 To required side AC 759.3 ---- 2.88043

Third work in triangle ACD to find angles D & C As sum of sides AD & AC 1256 ---- 6.50101 Is to their difference AD - AC 262.5 ---- 2.41929 So is tangent of 1/2 sum unknown angles D & C 58.15 ---- 10.20844 To tangent of 1/2 their difference D & C 18° 40' ---- 9.52874

Example 3rd Continued

Half difference two unknown angles C & D 18° 40' Half sum " " " C & D 58.15 Equal angle D 76.55

Half sums two unknown angles C & D 58° 15' Half difference " " C & D 18.40 Equal angle C 39.35

Fourth work in triangle ACD to find side DC requires As sine L C 39° 35' ---- 0.19572 Is to sine L A 63° 30' ---- 9.95175 So is given side AD 496.8 ---- 2.65609 To required side DC 697.6 ---- 2.84360

Example 4th Observing three steeples A B & C in a town at a distance whose distances asunder are known to be as follow viz AB 213 AC 404 and BC 262 yards. I took their angles of position from the place D where I stood which was nearest the steeple B and found the angle ADB 13° 30' and the angle BDC 29° 50'. Required my distance from each of the three steeples Answer AD 571 yard BD 385 yds & CD 514 yds