1. Right-angled triangles may, as well as others, be solved by means of the rule to the respective case under which any specified example falls and it will then be found, since a right angle is always one of the data, that the rule usually becomes simplified in its application.

2. When two of the sides are given, the third may be found by means of the property in Plane Geom. Triangles, prop. 16.

Hypoth. (base + perp.)

Base (hyp. perp.)

·

=

(hyp.+perp.). (hyp. - perp.) Perp. (hyp.- base) = √(hyp.+base).. (hyp.— base.) 3. There is another method for right angled triangles, known by the phrase making any side radius; which is this.

"To find a side.-Call any one of the sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write those words upon them accordingly. Call the word written upon each side the name of each side; then say,

As the name of the given side,

Is to the given side;

So is the name of the required side,
To the required side."

"To find an angle.-Call either of the given sides radius, and write upon it the word radius; observe whether the other sides becomes sines, tangents, or secants, and write those words on them accordingly. Call the word written upon each side the name of that side. Then say,

As the side made radius,

Is to radius ;

So is the other given side,

To the name of that side,

which determines the opposite angle."

4. When the numbers which measure the sides of the triangle are either under 12, or resolvable into factors which are each less than 12, the solution may be obtained, conformably

with this rule, easier without logarithms than with them. For,

Let A B C be a right angled triangle, in which a в, the base, is assumed to be radius; B c is the tangent

of A, and A c its secant, to that radius; or di-
viding each of these by the base, we shall
have the tangent and secant of A, respec-
tively, to radius 1. Tracing in like manner
the consequences of assuming в C, and A C,
each for radius, we shall readily obtain these expressions.

B

sec angle at vertex.

5. perpsin angle at base.
hyp.
base

6. =sin angle at vertex.
hyp.

SECTION II. On the Heights and Distances of Objects.

The instruments employed to measure angles are quadrants, sextants, theodolites, &c., the use of either of which may be sooner learnt from an examination of the instruments themselves than from any description independently of them. For military men and for civil engineers, a good pocket sextant, and an accurate micrometer (such as Cavallo's) attached to a telescope, are highly useful. For measuring small distances, as bases, 50 feet and 100 feet chains, and a portable box of graduated tape will be necessary.

We shall here present a selection of such examples as are most likely to occur.

EXAMPLE I.

In order to find the distance between two trees A and B, which could not be directly measured because of a pool which occupied much of the intermediate space, I measured the distance of each of them from a third object c, viz. Ac=588, BC=672, and then at the point c took the angle A C B between the two trees 55° 40'. Required their distance.

This is an example to case 2 of plane triangles, in which two sides, and the included angle, are given. The work, therefore, may exercise the student: the answer is 593.8.

EXAMPLE II.

Wanting to know the distance between two inaccessible objects, which lay in a direct line from the bottom of a tower on whose top I stood, I took the angles of depression of the two objects, viz. of the most remote 2540, of the nearest 57°. What is the distance between them, the height of the tower being 120 feet?

The figure being constructed, as in the margin, A B=120 feet, the altitude of the tower, and AH the horizontal line drawn through its top; there are given,

A

B

C

H A D=25° 30', hence B A D=B A H — H A D=64° 30'.
HA C=57° 0', hence B A C =BA H HA C=33° 0'.

H

Hence the following calculation, by means of the natural tangents. For, if A в be regarded as radius, в D and B C will be the tangents of the respective angles B A D, B A c, and c d the difference of those tangents. It is, therefore, equal to the product of the difference of the natural tangents of those angles into the height A B.

Thus, nat. tan 644°=2.0965436
nat. tan 33° 0.6494076

Standing at a measurable distance on a horizontal plane, from the bottom of a tower, I took the angle of elevation of the top; it is required from thence to determine the height of the

tower.

In this case there would be given A B and the angle A (see the figure in Right-angled Triangles), to find в C=A BX tan A. By logarithms, when the numbers are large, it will be, log. BC=log. A B+log. tan A.

Note.-If angle A=11°19' then B C A B very nearly.

A 16 42

A=21 48.
A=26 34

A=30 58

A=35 0

A=38 40
A=45°

3

AB

B C=

2

A B

To save the time of computation, therefore, the observer may set the instrument to one of these angles, and advance or recede, till it accords with the angle of elevation of the object; its height above the horizontal level of the observer's eye will at once be known, by taking the appropriate fraction of the distance ▲ B.

EXAMPLE IV.

Wanting to know the height of a church steeple, to the bottom of which I could not measure on account of a high wall between me and the church, I fixed upon two stations at the distance of 93 feet from each other, on a horizontal line from the bottom of the steeple, and at each of them took the angle of elevation of the top of the steeple, that is, at the nearest station 55° 54', at the other 33° 20'. Required the height of the steeple.

Recurring to the figure of Example II., we have given the distance c D, and the angles of elevation at c and D. The quickest operation is by means of the natural tangents, and the

Wishing to know the height of an obelisk standing at the top of a regularly sloping hill, I first measured from its bottom a distance of 36 feet, and there found the angle formed by

the inclined plane and a line from the centre of the instrument to the top of the obelisk 41°; but after measuring on downward in the same sloping direction 54 feet farther, I found the angle formed in like manner to be only 23° 45'. What was the height of the obelisk, and what the angle made by the sloping ground with the horizon?

The figure being constructed as in the margin, there are given in the triangle A C B, all the angles and the side A B, to find B C. It will be obtained by this proportion, as sin c (=17° 15'-B -A): A B (=54) :: sin A (=23° 45') : B c=73.3392. Then, in the triangle D B C are known в C as above, B D = 36, C B D = = 41°; to find the other angles, and the side c D. first, as C B B D C B

(D+c) (139°): tan

Thus,
B D :: tan

H

A

R

E

Hence 69° 30' + 42° 24' =

=

42° 24'=26° 54' BC D. Then, sin B C D B D :: sin C B D C D =51.86, height of the obelisk.

The angle of inclination D A E = HDA = C D B 22° 54'.

90° =

Remark.—If the line B D cannot be measured, then the angle DAE of the sloping ground must be taken, as well as the angles CA B and C B D. In that case D A E +90° will be equal to C D B: so that, after c в is found from the triangle A c B, C D may be found in the triangle C B D, by means of the relation between sides and the sines of their opposite angles.

EXAMPLE VI.

Being on a horizontal plane, and wanting to ascertain the height of a tower standing on the top of an inaccessible hill, I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°, then measuring in a direct line 180 feet farther from the hill, I took in the same vertical plane the angle of elevation of the top of the tower 33° 45'.. Required from hence the height of the tower.

The figure being constructed, as in the margin, there are given A B=180, C AB=33° 45', CAE 17° 15', C B D =

АСВ =СВЕ

=

=

11°, BDC=180°— (90° — D B E)=130°. And CD may be found from the expression C D rad AB sin A sin C B D cosec A C B sec D B E.