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and we have tang (

BC=) tang 15° 20' x tang 25° 58


° ' '

But log tang 25° 58' 9.687540
log tang 15 20 9.438059

log radius

log tang


2 -C

2 The angles B and C, and the side BC will be found as before.

EXAMPLE 2. Given the two sides of an obtuse-angled plane triangle respectively 1280 and 1860, and the included angle=270 45' ; required the remaining angles and side.

Ansner, the angles are 1/2° 54' 50",

and 39° 20' 10", and the side 940.237. EXAMPLE 3. In an acute-angled plane triangle, given the two sides AC, BC, respectively 120 and 112 yards, and the included angle C357° 58' 39"; what are the angles A, B, and side AB?

Answer, angle A=57° 27', B=64° 34' 21",

and the side AB=112.653.

whence Buon

=70 36



2 BC

2 a

AB (6): BDO*+62d

43. Given the three sides of a triangle, required the three angles.

Let BC=a,...AB=b ...AC=d...R=1...
and draw AD perpendicular on BC.
By prop. 47, Book 1, Euclid's Elements

Whence BD-
AB? + BC-AC2

a? +62-d

or BD-
Again in the right-angled triangle BDA, we have

-) :; rad (1): cos B=
2 a

2 ab de-a2+2 abb*_d?-(a+b) – 2 sind B(18) Therefore 1-cos Ba2-?

2 ab

2 ab From this expression we may deduce the following proportion : dato dtab Х

:: R? : sinB; which, calling a the 2

2 semiper meter, becomes ab: (9-a) (946):: R?: sin’ | B; and by this expressinn we usually calculate the value of an angle from the three sider

We may also calculate it by means of the formula cos B R

(a? +62-d”), which gives 2 ab

2 ab ; a' +62-d*::R: cos B.

b B


ab :




That is, to find an angle B in a triangle, of which we know the three sides, we must first find the sum of the squares of the two sides containing the angle required, and subtract from this sum the square of the third side; this will leave a remainder. We must then construct this proportion. The double of the rectangle of the side containing the angle required is to this remainder, as radius to the cosine of the angle required. If the angle B is obtuse, we shall have

cos B= (d_a?–64)

2 ab 44. A formula equally convenient for logarithmic calculation, may be obtained as follows: We have just shewn that 1-cos B=2 sin

B_2.(9-a) 9–6)

ab And 1 + cos B=1+

a2 +62-od? _ (a+b)2--d2 _ (a+b+d)(a+bood
2 ab
2 ab

2 ab 2. (

B =?

by (18); consequently we have ab

2 sin' B

(9-a) (946)

= tang?


(9-a)(9-6 cog B

9 (940); (9) (9-0) 2

ab if radius=1; or BR (q-a) (9–6), if radius=R

9 (9-0) N. B. This formula must not be used, when the angle sought for is nearly 180°

-2-969–d) –2 coo




Examples. 1. In a plane triangle given the three sides respectively 1756, 1214, 1992, required the angles. By the first formula we have

B_R’(9a) (9–6), sin?

ab q representing the semiperimeter, and a=1756, b=1214, the sides about the angle B.


1756 +1214 +1392 = 2181 and 425, and qub=967


Here q= {


= 20.

log sin

But log R?

log (9-a)= 425 2.628389

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


16.755475 arith. com. log b

=1214 16.915781

B log sin 2

2 19.285071 B

9.642535 2

B whence the angle = 26° 3', very near.y, and therefore the angle

2 B=52° 6'. With equal facility the angle B may be founa from the formula

B_R* (-a) (946)

For log R

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

= 16.661344
= 789 =

log tang*

2 |19.378082

= 20.

arith. com. log 9
arith. com. log 9-d

- 2181

log tang

whence Z



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

Answer, 37° 59' 53"

52° 37' 46"
89° 22' 40"

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, x 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::*:y. Therefore ay =

y? sin A

=2 s. Hence

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cos A

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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, 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. If the quadrant is moved

B 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 right-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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