In the triangle BAD, to find the angle BAD.

=

Hence, 90° - DAC = 90° – 51° 34′ 40′′ 38° 25′ 20′′ = C, and, 90° — BAD= 90° — 32° 18′ 35′′ – 57° 41′ 25′′. and, BAD+ DAC-51° 34' 40" + 32° 18′ 35′′ 83° 53′

=

==

B,

2. In a triangle, of which the sides are 4, 5, and 6, what are the angles?

Ans. 41° 24' 35"; 55° 46′ 16"; and 82° 49' 09".

SOLUTION OF RIGHT-ANGLED TRIANGLES.

34. The unknown parts of a right-angled triangle may be found by either of the four last cases; or, if two of the sides are given, by means of the property that the square of the hypothenuse is equivalent to the sum of the squares of the two other sides. Or the parts may be found by Theorems IV. and V.

2. In a right-angled triangle BAC, there are given AC=384, and B= 53° 08′: required the remaining parts. Ans. AB=287.96; BC= 479.979; C 36° 52'.

=

APPLICATION TO HEIGHTS AND DISTANCES.

I. To determine the horizontal distance to a point which is inaccessible by reason of an intervening river.*

35. Let C be the point. Measure

along the bank of the river a hori

zontal base line AB, and select the
stations A and B, in such a man-
ner that each can be seen from the
other, and the point C from both
of them. Then measure the hori-
zontal angles CAB and CBA, with
an instrument adapted to that purpose.

A

B

Let us suppose that we have found AB=600 yards, CAB= 57° 35', and CBA = 64° 51′.

The angle C180° - (A+B) = 57° 34'.

II. To determine the altitude of an inaccessible object above a

Measure any horizontal base line, as BA; and at the extremities B and A, measure the horizontal angles CBA and CAB. Measure also the angle of elevation DBC.

Then in the triangle CBA there will be known, two angles and the side AB; the side BO can therefore be determined. Having found BC, we shall have, in the right-angled triangle DBC, the base BC and the angle at the base, to find the perpendicular DC, which measures the altitude of the point D above the horizontal plane BC. Let us suppose that we have found

BA=780 yards, the horizontal angle CBA = 41° 24′; he horizontal angle CAB=96° 28', and the angle of eleva tion DBC= 10°43'.

In the triangle BCA, to find the horizontal distance BC The angle BCA = 180° — (41° 24′ +96° 28′) = 42° 08′ = C.

In the right-angled triangle DBC, to find DC.

REMARK I. It might, at first, appear, that the solution which we have given, requires that the points B and A should be in the same horizontal plane; but it is entirely independent of such a supposition.

For, the horizontal distance, which is represented by BA, is the same, whether the station A is on the same level with B, above it, or below it. The horizontal angles CAB and CBA are also the same, so long as the point C is in the vertical line DO. Therefore, if the horizontal line through A should cut the vertical line DC, at any point, as E, above or below C, AB would still be the hori zontal distance between B and A, and AE, which is equal to AC, would be the horizontal distance between A and C.

If at A, we measure the angle of elevation of the point D, we shall know in the right-angled triangle DAE, the base AE, and the angle at the base; from which the perpendicular DE can be determined.

37. Let us suppose that we had measured the angle or elevation DAE, and found it equal to 20° 15'.

First: In the triangle BAC, to find AC or its equal AE.

In the right-angled triangle DAE, to find DE.

which expresses the vertical distance that the station B is above the station A.

REMARK II. It should be remembered, that the vertical distance which is obtained by the calculation, is estimated from a horizontal line passing through the eye at the time of observation. Hence, the height of the instrument is to be added, in order to obtain the true result.

SECOND METHOD.

38. When the nature of the ground will admit of it, measure a base line AB in the direction of the object D. Then measure with the instrument the angles of elevation at A and B.

Then, since the outward angle DBC is equal to the sum of the angles A and ADB, it follows that the an

D

gle ADB is equal to the difference of the angles of elevation at A and B. Hence, we can find all the parts of the triangle ADB. Having found DB, and knowing the angle DBC, we can find the altitude DC.

This method supposes that the stations A and B are on the same horizontal plane; and therefore it can only be used when the line AB is nearly horizontal.

Let us suppose that we have measured the base line, and the two angles of elevation, and