being in the dusk of evening, with my telescope, I descried the lamp of the light-house in the horizon, at which time my eye was elevated 6 feet above the surface of the water. What is the height of the light-house above the water?

OPERATION.

5 leagues 16.5 miles.

Then 16.5-(42056462 +6, × 6)=13.943 miles, equal 73619 feet, nearly.

Then, 73619×73619,÷42056462=129 feet, nearly, Ans. PROB. III. To determine a right-angled triangle, whose base coincides with the diameter of a given semi-circle, about which it is circumscribed; the ratio of the hypothenuse and perpendicular being given.

RULE. Take any two numbers in the same ratio of the hypothenuse and perpendicular; then, as the hypoth. is to sine 90, so is the perpend. to sine of the angle opposite. And as radius given to the semi-circle is to sine 90, so is tang. double the angle first found to the actual length of the perpendicular, from which, by means of the given proportion, the other sides may be found.

PROB. IV.-To Measure a Triangle.

RULE. From half the sum of the three sides, subtract each side severally; multiply these three remainders and the said half sum continually together; the square root of the last product will be the area of the triangle.

PROB. V.-To Measure any irregular plane figure.

RULE. The whole may be divided into triangles, and measured separately; the sum of the area of the triangles will be the area of the whole.

PROB. VI.-Given, the minutes of a Survey, to draw a Map of it, and to find the area by Construction.

RULE.-Draw a line to represent a Meridian, from which lay off a Bearing or Course of the first side of the field, from a line of Chords; and from a scale of equal parts measure the length of the side, and draw a line to represent it; at the end of this line, draw a line parallel to the meridian line, and then lay off the second side of the field, as before taught ; proceed in the same manner to draw parallel lines, and to lay off the several sides, till the whole is protracted; divide this map into tri

angles, by drawing diagonals, and the sum of the areas of these triangles is the area of the field.

NOTE.-In protracting a survey, let the top of the paper be considered as N., the bottom S., the right hand as E., and the left hand W.; lay the course to the right or left of the meridian line, according as it is E. or W., and from the upper or lower part of the line, according as is N. or S.

-N.B.-To find the area of a field, by calculation, the sides and angles of oblique-angled triangles, belongs, more properly to what is contained in some work on Surveying, and consequently for this and other reasons will be omitted in this.

PROB. VII.-Given, the Segment of a Circle, to find the Length of the Arch.

RULE.-Divide the segment into two equal parts; then measure the chord of the half arch, from the double of which subtract the chord of the whole segment; and one-third of that difference being added to the double of the chord of the half arch, will give the length of the arch line.

EXAMPLE.

The whole chord of a segment is 216, either of the other sides is 126. Required, the arch line.

* 126×2,-216÷3,+252-264, length of the arch line. PROB. VIII.-Given, the Chord and versed Sine of a Segment, to find the Diameter of a Circle.

RULE.-Multiply half the chord by itself, and divide the product by the versed sine; add the quotient to the versed sine; the sum will be the diameter.

N.B.-The diameter of a circle to its circumference, is as 1 to 3.14159265358979323846264338387950288419716939937 510582097494 4592307816405286288998628034 8253421170 679, nearly. Decimals 100 in number.

To find a square that is exactly equal to a given circle, has never yet been discovered, and probably never will.

PROB. IX. To find the length and breadth of a parallelogram, when the area is given, and the length exceeds the breadth by a certain number of rods.

RULE.-Square half the number of rods that the length ex

ceeds the breadth, add the square to the area, and to the square root of the sum add the half number of rods which the length exceeds the breadth; the sum is the length; subtract the difference of the two sides from the length; this difference is the breadth.

PROB. X.-The diagonal and length of a parallelogram being given in one sum, and the breadth separately, to find the length and area.

RULE. Divide the square of the breadth by the sum of the diagonal and length; the quotient is the excess of the diagonal above the length, half of which excess, subtracted from half the sum of the diagonal and length, will leave the length, which multiplied by the breadth, gives the area.

PROB. XI. The diagonal and the breadth of a parallelogram being given in one sum, and the length separately, to find the breadth.

RULE. From the square of the diagonal and breadth, subtract the square of the length, divide the remainder by twice the sum of the diagonal and breadth, the quotient will be the breadth.

PROB. XII.-Given, the diagonal and area of a parallelogram, to find the length and breadth.

RULE. 1.-Divide the area by the diagonal, and square the quotient. Square half the diagonal, and the square root of the difference of the squares, subtract from half the diagonal.

2. Square the remainder, and add the square of the quotient, and the square root of the sum of the two squares will be the breadth, by which divide the area, the quotient is the length.

PROB. XIII.—To measure a Cylinderoid, or Prismoid.

RULE. To the areas of both bases, add a mean area, that is, the square root of the product of the two bases, multiply that sum by a third of the height or length, the product is the solidity.

PROB. XIV. To find how large a Cube may be cut from any given sphere, or be inscribed in it.

RULE.-Extract the square root of one-third the square of the diameter, the root is the side of the required cube.

PROMISCUOUS QUESTIONS IN MENSURATION.

1. What is the Hypothenuse and Perpendicular of a rightangled triangle, whose sides are in the ratio of 7 to 4, the base being 100?

*

7X7,-4X4-5.7445, ratio of the base.
5.7445:4: 100: 69.68, Perpendicular,

Then,{

5.7445:4::100: }

:7:"121.94, Hypothenuse,

Ans.

2. In an olique-angled triangle, the product of the two sides is 186, and their difference 3.5, the shortest side is to the base in the ratio of 4 to 7. Required, the sides.

* 3.5 X 3.5, 4-3.0625, +186-189.0625; its square root is 13.75.

Then, 13.75-3.5÷2=12, shortest side.
And, 13.75+1.75-15.5, longer side.

And, if4:12::7:21, base.

3. In a right-angled triangle, the difference of the sides is 70 rods, and the difference of the segments of the hypothenuse, made by a perpendicular let fall from the angle opposite, is 98. Required the sides.

* In questions of this nature, the difference of sides will form the hypothenuse, and the difference of the segments the sum of the two legs of a triangle, in all respects similar to the given one.

Therefore, 98X 98,-70X70-4704.

70X70-4704-14, difference of sides.

And, 98+14,÷2=56; also, 98-14,÷÷2-42.

Then, as

14:70:56:280, Base,

14:70::42:210, Perpendicular, Ans.
14:70:70:350, Hypothenuse,

4. In a triangle, the area is 216, the angle at the base 36° 52', and the cube of the sides is 46656. Required, the sides. * The base, multiplied by the perpendicular, must be 432. As the sine of the given angle is to 3, so is the co-sine of the same angle to 4, calling the base 4, and the perpendicular 3; and 4X4-16; also, 3×3=9; and 4×3=12, twice the

area.

12:16:432:576,576-24, Base of Area,

Then, as 12: 9::432:324,324-18, Perpendicular

(24X24+18X18)=30, Hypothenuse,

Ans.

5. There are two columns in the ruins of Persepolis, left standing upright; one is 70 feet above the plane, and the other 50 ft.; in a straight line between these, stands a statue, 5 feet in height; the head of which is 100 feet from the summit of the higher, and 80 feet from the top of the lower column. Required, the distance between the tops of the two columns.

* 100 × 100,--70-5X70-5-5775; /5775-75.99342. 4375-66.143782.

80X80,-50-5X50-5-4375;

75.9934266.143782220202.984192388804.

70-50-20; 20×20=400.

(20202.984192388804+400)=143.5373396, feet, Ans.

6. Admit 10 hhds. of water are discharged through a leaden pipe, of 2 inches diameter, in a certain time. What must be the diameter of another pipe that shall discharge four times as much in the same time?

* 21×2}=61; 61=6.25, and 4 is the given proportion. 6.25X4-25; then, /25-5 inches, Ans.

7. Two wheels of unequal dimensions, made fast to an axle 11 feet long, are set to rolling on an even plain; the path described by the less wheel, encloses just 314.16 square rods; and the less wheel is 5 feet in diameter. Required, the diameter of the greater.

*(314.16.7854)=20,X16-330 feet, diameter of the 314.16 square rods; 11×2=22.

Then, 330+22=352 feet, diameter of the circle formed by the rolling of the large wheel.

Then, as 330 feet: 5 feet: :352 feet: 5 feet, diameter of the greater wheel, Ans.

8. In turning a chaise within a ring of a certain diameter, I observed the outer wheel to make two turns, and the inner wheel to make but one; both wheels were four feet high. If both were fixed to an axle 5 feet asunder, what was the circumference described by the outer wheel?

* By the question, the outer ring must be twice the diameter of the inner ring. The distance between the rings being 5 feet, it follows that the diameter of the inner ring will be 10 feet, and the diameter of the outer ring 20 feet.

Then, if 113:355::20:62.83, &c. feet, Ans.

9. An orchard of 2400 mulberry trees were arranged that the