CHAPTER VII.

SOLUTION OF TRIANGLES.

107. Triangles. — In every triangle there are six elements, the three sides and the three angles. When any three elements are given, one at least of the three being a side, the other three can be calculated. The process of determining the unknown elements from the known is called the solution of triangles.

NOTE. - If the three angles only of a triangle are given, it is impossible to determine the sides, for there is an infinite number of triangles that are equiangular to one another.

Triangles are divided in Trigonometry into right and oblique. We shall commence with right triangles, and shall suppose C the right angle.

RIGHT TRIANGLES.

108. There are Four Cases of Right Triangles.

I. Given one side and the hypotenuse.
II. Given an acute angle and the hypotenuse,
III. Given one side and an acute angle.
IV. Given the two sides.

B Let ABC be a triangle, right-angled at C, and let a, b, and c, as before, be the sides opposite the angles A, B, and C, respectively.

b

с The formulæ for the solution of right triangles are (1), (2), (3) of Art. 94.

A

109. Case I. Given a side and the hypotenuse, as a and c; to find A, B, b. We have

sin A

a

с

.. log sin A = log a - log c, from which A is determined ; then B= 90° – A. Lastly,

b=

= C COS A. .. log b= log c + log cos A. Thus A, B, and b are determined.

Ex. 1. Given a= 536, c= 941; find A, B, b.

=

Logarithmic Solution. log sin A = log a – log c.

log b= log c + log cos A. log a = 2.7291648

log c= - 2.9735896 log c= 2.9735896 log cos A 9.9148283 log sin A * = 9.7555752

log b= 2.8884179 .. A= 34° 43' 22".

.. b = 773.424.

=

.:. B= 55° 16' 38".

Our two methods of calculation give results which do not quite agree. The discrepancies arise from the defects of the tables.

The process of solution by natural sines, cosines, etc., can be used to advantage only in cases in which the measures of the sides are small numbers.

We might have determined b thus:

b= V(c - a)(c + a);

=

or thus:

b= a tan B.

NOTE. — It is generally better to compute all the required parts from the given ones, so that if an error is made in determining one part, that error will not affect the computation of the other parts.

To test the accuracy of the work, compute the same parts by different formulæ.

Ex. 2. Given a=21, c= 29; find A, B, b.
Ans. A
46° 23' 50",

B

43° 36' 10",

NOTE. - In these examples the student must find the necessary logarithms from the tables.

110. Case II. - Given an acute angle and the hypotenuse, as A and c; to find B, a, b.

Also

a=csin A, and b=ccos A.

... log a= log c + log sin A; and

log b= log c + log cos A. Thus B, a, and b are determined.

We have

a=csin A, and b=ccos A. Using a table of natural sines, we have a= 125 x .813778

101.722, and

b=125 X.581177 72.647.

Logarithmic Solution. log a= log c + log sin A. log b = log c + log cos A. log c = 2.0969100

log c= - 2.0969100 log sin A : 9.9105057

log cos A = 9.7643080 log a * = 2.0074157

log b* = 1.8612180 .. a = 101.722.

.b= 72.647.

Ex. 2. Given A = 37° 10', c=8762; find a and b.

Ans. 5293.4; 6982.3.

111. Case III. Given a side and an acute angle, as A and a; to find B, b, c.

B= 90° — A= 57° 44' 36''. log b= log a - log tan A. log c= log a – log sin A. log a = 3.7381858

log a = 3.7381858 log tan A = 9.8001090

log sin A = 9.7273076 log b = 3.9380768

log c= 4.0108782 ..b= 8671.152.

.: C= 10253.64.

=

Ex. 2. Given A = 34° 18', a = 237.6; find B, b, c.

Ans. B = 55° 42'; b= 348.31; c=421.63. 112. Case IV. Given the two sides, as a and b; to find A, B, C.

a We have tan A

I ;

then B = 90° – A. b

=

a

=

Also

C= a cosec A

sin A .:: log tan A = log a – log b, and

log c = log a – log sin A. Ex. Given a= 2266.35, b= 5439.24; find A, B, C.

Solution. log tan A = log a – log b. log c= log a – log sin A. log a= 3.3553270

log a = 3.3553270 log b= 3.7355382 log sin A = 9.5850266 log tan A = 9.6197888

log c= 3.7703004 .:. A = 22° 37' 12".

..c= 5892.51. ... B= 67° 22' 48'.

=

NOTE. — In this example we might have found by means of the formula C=Va? + b2; but we would have had to go through the process of squaring the values of a and b. If these values are simple numbers, it is often easier to find c in this way; but this value of c is not adapted to logarithms. A formula which consists entirely of factors is always preferred to one which consists of terms, when any of those terms contain any power of the quantities involved.

113. When a Side and the Hypotenuse are nearly Equal.

When a side and the hypotenuse are given, as a and c in Case I., and are nearly equal in value, the angle A is very near 90°, and cannot be determined with much accuracy from the tables, because the sines of angles near 90° differ very little from one another (Art. 81). It is therefore desirable, in this case, to find B first, by either of the following formulæ :

B 1 sin

(Art. 50) 2

2

cos B