log sin



arith. com. log 2

But log R

log (9-a)= 425 2.628389

log (9--6)= 967 2.985426 arith. com, log a


16.755475 arith. com. log b

=1214 = 16.915781

log sin?

2 19.285071


B whence the angle =26° 3', very nearıy, and therefore the angle B-52° 6'. With equal facility the angle B may be founa from the formula

B tang

R* (-a) (9-6)

9 (9-0)
For log Ro

log (9-a)= 425 = 2.628389
log (9–6)= 967 = 2.985426

—2181 = 16.661344 arith. com. log 9-d

= 789 = 17.102923

B В.
log tang

2 19.378082


= 26° 3', very nearly, and therefore B=52° 6', as before,

In a similar manner the other angles may be calculated, and will be found 43° 28', and 84° 26'.

EXAMPLE 2. In a plane triangle suppose the sides are 510, 760, and 384, what are the angles ?

Answer, 27° 4' 32",

115° 43' 50,

37° 11' 38". EXAMPLE 3. Suppose that the three sides of a triangle are 112.65, 120, and 112; required the angles.

Answer, 57° 58' 32",

57° 27' 3",

64° 34" 25". EXAMPLE 4. Given the three sides of a plane triangle 33, 42.6, and 53.6; what are the angles ?

Answer, 37° 59' 53"

52° 37' 46" 89° 22' 20".&

log tang 2

whence Love


45. Such are the principles of plane trigonometry. Other problems concerning triangles might be proposed; but not to multiply examples, we shall add only the following one:

Given one of the acute angles of a right-angled plane triangle, together with its surface ; required the three sides.

Let A be the given angle, the opposite side, y the other side, and s the surface of the triangle. By a well known theorem wy= 2 s; and again, sin A : cos A::x : y. Therefore ay =

yz sin A

-2 s. Hence

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One of the most direct and useful applications of Plane Trigonometry is to the mensuration of heights and distances.

For this purpose the observer must be provided with a quadrant for measuring vertical angles, and a theodolite for taking angles in a horizontal direction. He must also have a Gunter's chain, a set of 50 foot tapes, the common measuring rods, or some equivalent instruments for ascertaining distances by actual measurement.

The use of these instruments is best acquired under the direction of a person practically acquainted with their adjustment and application. I shall however proceed to give some idea of the mode in which they are employed. I. In the adjoining diagram,

A CDE is a quadrant, or quarter-circle; C its centre, A any object elevated above the eye of the observer, CB a line parallel to the horizon, and CW a weight or plumb-line hanging freely from C, and therefore perpendicular to CB.

-B If the quadrant is moved round C, till the object at A

E is visible through the two sights a, b; then the arc EF will measure the angular distance of the object A above the horizon. For the angles BCW, and ACE are rig!it-anglet, (since _ ACE=L DCE, and since the weight is perpendicular to the horizon); take away the common angle BCE, and the remaining angle ECF is equal to the remaining angle ACB. Therefore the arc EF which measures the < ECF, determines the number of degrees, minutes, &c. of the angle ACB.

II. The theodolite DCF is a graduated circular instrument with two indices moveable round the center C; A and B are two objects


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47. To find the height of an accessible object, as AR.

Recede from the object AE along the line ED, to some convenient distance, which measure. With the quadrant take the angle ACB; then in the triangle ACB, there is given the side BC, and the angle ACB; hence the side AB may be found by B (38 Pl. Trig.) and therefore the height required.

(N. B. BE or CD represents the E height of the eye of the observer above the horizontal plane.)


Example. Let the horizontal distance ED be 200 feet, and the angle of elevation ACB=37° 35'; supposing the height of the eye to be 5 feet ; required the height of the tower. By (38 Pl Trig.) As rad. ; tang ACB::ED or CB : BA. Neg. log radius

iog tang ACB (37° 35') 9.886288
log CB (200)

log AB=153.92

2.187318 Whence, adding 5 feet for the height of the observer's eye above the horizon, we have AE the height of the tower=158.92 feet.

= 1o.




48. To find the height of an inaccessible object, u CD.

At any two convenient stations A, B in the same vertical plane

1) with CD, observe the angles of elevation DAC, DBC, and measure the distance AB or GE Then because the exterior angle DBC is equal to the two interior angles


в. BDA, DAB; if DAB be subtracted from DBC, the angle BDA will remain. Therefore in the triangle ADB we shall have the side AB, and the angles DAB, ADB, whence the side DB will be found by Case I. Plane Trig. And then in the right-angled triangle DBC, we shall have the angle DBC, and the side ÖB just found, whence DC may be determined, and consequently DF, the height required.

Example. In the above figure let the angle of elevation DBC be 48°; and at another station in the same vertical plane, but 200 feet farther off in the same direction, let the angle of elevation be 26° 45', the height of the eye being 5 feet ; required the height of the tower.

First _ ADB=48°—26° 45=219 15'.

Then to find BD arith. com. log sin ADB (21° 15')

10.440766 log sin DAB (26° 45')

= 9.653308 log AB (200)

2.301030 log BD=248.87

And to find CD
Neg. log radius

= io.
log sin DBC (48) 9.871073
log BD (248.87)

log CD=184.57

2.266177 Whence DF=189.57 feet, the height of the tower.

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49. To find the distancc of an inaccessible object. Let C be any inaccessible object, and

C A, B two points from which the distance of that object is to be found. Measure the distance AB, and observe the horizontal angles BAC, ABC. Then in the triangle ABC, we have given the base AB, and the


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