Page images
PDF
EPUB

(10) Given a = 396, b=403, log a = 2·5976952,

log b = 2.6053050,

I tan 44°. 29′ = 9.9921670, Z tan 44o. 30′ = 9·9924197.

194. We shall now give a few Problems to illustrate the practical use of the methods of solution of triangles explained in this Chapter.

EXAMPLES.-LI.

(1) Having measured a distance of 220 feet in a direct horizontal line from the bottom of a steeple, the angle of elevation of its top was found to be 46°. 30'. Required the height of the steeple.

Given log 220-2-3424227, log tan 46°. 30' 10.0227500,

log 2.318353651727.

=

(2) A river AC whose breadth is 200 feet runs at the foot of a tower CB, which subtends an angle BAC of 25°. 10' at the edge of the bank.

Required the height of the tower, given

log 56989700, L tan 25°. 10' 9.6719628,

=

[blocks in formation]

(3) A person on the top of a tower, whose height is 50 feet, observes the angles of depression of two objects on the horizontal plane which are in the same straight line with the tower to be 30° and 45o. Find their distances from each other and from the observer.

(4) At 140 feet from the base of a tower, and on a level with the base the angle of elevation of the top was found to be 54o. 7′. Find the height of the tower, having given

[blocks in formation]

(5) A person observes the angle of elevation of a hill to be 32o. 14′, and on approaching 500 yards nearer, he observes it to be 63°. 26'. Find the height of the hill, having given

tan 32°. 14′ = 63, tan 63o. 26′ = 1·998.

(6) A tower 150 feet high throws a shadow 75 feet long upon the horizontal plane on which it stands. Find the sun's altitude, having given log 2 = 3010300,

I tan 63°. 26' = 10.3009994, Z tan 63o. 27′ = 10·3013153.

(7) A tower stands by a river.

A person on the opposite

bank finds its elevation to be 60°: he recedes 40 yards in a direct line from the tower, and then finds the elevation to be 50°. the breadth of the river, having given tan 50° 1.19.

=

Find

(8) A rope is fastened to the top of a building 60 feet high. The length of the rope is 109 feet. Find the angle at which it is inclined to the horizon.

Given

sin 33. 23′ = '5502, sin 33. 24 = 55048.

(9) A tower is 140 feet in height. At what angle must a rope be inclined to the horizon which reaches from the top of the tower to the ground, and is 222 feet in length ?

Given sin 39°. 18′ = •6333809, sin 39. 19′ = 6336059.

(10) A person standing at the edge of a river observes that the top of a tower on the edge of the opposite side subtends an angle of 55° with a line drawn from his eye parallel to the horizon; receding backwards 30 feet, he then finds it to subtend an angle of 48°. Find the breadth of the river.

Given

L sin 7° 9.08589, L sin 35° = 9.75859,

I sin 48° 9:87107, log 3=47712, log 1.0493 = '02089.

=

S. T.

11

(11) Standing straight in front of the corner of a house which is 150 feet long, I observe that the length subtends an angle

1

whose cosine is and its height subtends an angle whose sine

[blocks in formation]
[ocr errors]

; determine the height.

(12) Standing straight in front of one corner of a house, I find that its length subtends an angle whose tangent is 2, while its height subtends an angle whose tangent is the house is 45 feet, find its length.

3 5

: the height of

CHAPTER XVI.

ON THE SOLUTION OF TRIANGLES OTHER THAN RIGHT-ANGLED.

195. In the solution of triangles other than right-angled we meet with four distinct cases, the following being the data.

(1) The three sides a, b, c.

(2) Two angles and a side, as A, C, b.

(3) Two sides and the angle between them, as a, b, C. (4) Two sides and an angle opposite one of them, as a, b, A. These cases we shall discuss in the four following articles.

196. Case I. When a, b, c are given, we have the formula established in Art. 176,

cos A

=

b2 + c2-a2
2bc

from which A can be found.

Or we can find A from the formula established in Art. 181,

[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

and

197.

First,

[ocr errors]

from which C can be found,

B = 180° - (A + C), from which B can be found.

Case II. When A, C, b are given.

B = 180° − (A + C), from which we can find B.

[blocks in formation]

sin B'

Next,

=

b

C

sin C

Lastly,

[blocks in formation]

198. Case III.

from which we can find a.

from which we can find c.

When a, b, C are given, we may take the

formula established in Art. 176,

c2 = a2 + b2 - 2ab. cos C, from which we can find c.

[ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors]

Lastly,

B

from which we can find A.

B = 180° - (A + C), from which we can find B.

Or we may proceed to find A and B before we find c, thus: by the formula established in Art. 177,

[blocks in formation]
« PreviousContinue »