let fall the perpendicular CD; and that will be the height of the object required.

The external angle CBD, of the triangle ABC; is equal to the two internal opposite ones, A, and ACB (by theo. 4. sect. 1.) wherefore if one of the internal opposite angles be taken from the external angle; the remainder will be the other internal opposite one, thus;

CBD 55°-A 37°-ACB 18°.

Therefore in the triangle ABC; we have the angles A, and ACB, with the side AB given; to find BC,

As S. ACB: AB :: S. A : BC,=169.4.

Having found BC, we have in the triangle BCD, the angle CBD 55° consequently BCD 35° and BC 169.4; to find DC.

This is performed by case the first, of right-angled trigonometry, three several ways; thus:

As R: BC: S. CBD : DC,=138.8,

The height required,

If BD, the breadth of the moat, were required; it may also be found, by three different statings, as in the first case of right-angled plane trigonometry.

Let BC, a may-pole, whose height is 100 feet, be broken at D; the upper part of which, DC, falls upon a horizontal plane, so that its extremity, C, is 34 feet from the bottom or foot of the pole.

Required, the segments BD and DC.

Lay 34 feet from A to B; on B erect the perpendicular BC of 100 feet; and draw AC: bisect AC (by prob. 4. sect. 1.) with the perpendicular line, EF; and from D, where it cuts the perpendicular BC, draw AD, which will be the upper segment; and DB will be the lower.

By cor. to lemma preceding theo. 7. sect. 1. AD =DC; and (by the lemma) the angle C-CAD.

In the triangle ABC, find C, as in case 6, of rightangled trigonometry, thus,

1. As BC : R :: AB : T. C=GAD,=18° 47'.

By theo. 4. sect. 1. The external angle ABD= 37° 34' or to twice the angle C, i. e. to C and GAD.

Then in the triangle ABD, there is ADB 37° 34′ therefore also its complement DAB 52° 26' and AB 34, given, to find AD and BD.

By the second case of rectangular trigonometry.

2. As S.ADB: AB::R: AD or DC,=55.77. Hence, BC-DC=BD,=44. 25, as required.

PROB. VII.

To take the altitude of a perpendicular object on a hill from a plane beneath it.

Let the height DC, of a wind-mill on a hill be required.

From any part of the plane whence the foot of the object can be seen, let angles be taken to the foot and top; measure thence any convenient distance towards the object; and at the end thereof, take another angle

On an indefinite blank line, lay the stationary distance AB 104 feet; from A, set off the second, and from B, the third given angle; and from the intersecting point C of the lines formed by them, let fall the perpendicular CE : from A set off the first angle, and the line formed by it will determine the point D. Thus have we the height of the hill, as well as that of the wind-mill.

The angle CBE—A=ACB, as in the last prob.

In the triangle ABC, find AC thus,

As S.ACB: AB :: S.ABC (or sup. of CBE): AC

=333.6

The angle CAE-DAE-CAD.

The angle ADC=AED+EAD, by theo. 4.

In the triangle CAD, find CD thus,

As S.ADC : AC :: S.CAD: DC=86.46 required. CE, BE, or DE, may be found by various other statings, as set forth in the first and second cases of rectangular trigonometry.

PROB. VIII.

To find the length of an object, that stands obliquely on the top of a hill, from a plane beneath.

Let CD be a tree whose length is required.

This is done at two stations.

Make a station at B, from whence take an angle to the foot, and another to the top of the tree; measure any convenient distance backward to A, from whence also let an angle be taken to the foot and another to the top; and you have the requisites given. Thus,

First station. Angle to the foot EBD=36° 30'. Angle to the top EBC=44° 30'.

Stationary distance AB=104 feet.

Second station. Angle to the foot EAD=24° 30'. Angle to the top EAC-32° 00'.

Let DC and DE be required.

The geometrical constructions of this and the next problem are omitted: as what has been already said,