Or, again, dividing (57) by (58), we find

in which r denotes the radical factor, or the radius of the inscribed circle of the triangle.

In the solution of right triangles, the formulas given directly by the definitions of the functions, (Art. 10), are employed. Thus, in

the triangle BAC right

angled at A we have

= sin B2

a

b

and

= tan B

с

Whence ba sin B = c tan B

=

Similarly, c = a sin C b tan C.

FIG. 4.

(60)

That is, Either perpendicular side of a right triangle

equals the hypotenuse multiplied by the sine, or the other

perpendicular side multiplied by the tangent of the angle opposite the first mentioned side.

Cor. 1 Since the angles B and C are the complements of each other, formulas (60) give

and

b = a cos C = c cot C)

c = a cos B = b cot B

(61)

That is, Either perpendicular side of a right triangle equals the hypotenuse multiplied by the cosine, or the other perpendicular side multiplied by the cotangent of the angle adjacent the first mentioned side.

Cor. 2. From (61), we obtain

That is, The hypotenuse of a right triangle equals either perpendicular side multiplied by the secant of the adjacent angle.

III. SOLUTION OF TRIANGLES.

I. OF RIGHT TRIANGLES.

24. 1. In the right triangle ABC, given the hypotenuse a = 255 ft., and the angle B = 57° 14′, to find the angle C and the sides b and c.

Solution.-1. The angle C = 90° · 57° 14' 32° 46'. 2. By (60), b = a sin B, whence, by logarithms, (Art. 5), a = 255 ft. log 2.406540

REMARK.-We drop 10 from the sum of the logarithms since the tabular log sin B is ten too large, (Art. 20).

3. By (61), c = a cos B, whence

α = 255 log 2.406540

2. In the right triangle ABC, given the side b=150 and the angle B = 40° 30', to find the angle C, the hypotenuse a, and the side c.

Solution.-1. The angle C 90°-40° 30' = 49° 30'.

3. In the right triangle ABC, given the hypotenuse b =243 and the side a = 120, to find the other parts.

A = 29° 35' 33'' log sin *9.693575

2. The angle C′ = 90°

29° 35' 33 = 60° 24' 27''.

с

a = 123

log 2.089905

log 4.649812

log 2.324906

c = 211.30

4. In a right triangle ABC, given the side b = the side c = 256, to find the other parts.

Solution.-1. By definition, tan C =

с

whence

b

*Since tabular logarithms are true logarithms plus 10.

385, and

25. CASE 1.—Given two sides and an opposite angle.

In the triangle ABC, given a = 60, b = 50, and A = 60° 40' 30'', to find the angles B and C and the side c

B = 46° 35' 51'' log sin 9.861263, (Art. 6, Sch.).

SCH.-Under the above case, in the solution of triangles there are

often two solutions--two values of each of the unknown parts, which àre consistent with the given ones. This arises from the fact that the angle required, opposite one of the given sides, is found from its sine, which is the same precisely, (17), for the possible value of the required angle and for its supplement. A full discussion of the case cannot conveniently be taken up here. We give simply its conclusion as enunciated in the following

Principle -If the side opposite the given angle is equal to or greater than the other given side, the required angle is acute and the triangle has but one solution; but if the side opposite the given angle is less than the other given side, the required angle may be acute or obtuse, right or imaginary, and the triangle has two, one, or no solution. The fact of a right or an imaginary angle is readily detected in the course of the solution.

26. CASE 2. Given two angles and an opposite side.

In the triangle ABC, given the angle A = 40° 15', the angle B = 35° 12' 8'' and the side a = 50 ft., to find the angle C and the sides b and c.

Solution.-1. The angle C=180°—(A+B)=104° 32′ 52''.

27. CASE 3. Given two angles and included side.

28. CASE 4. Given two sides and included angle.

In the triangle ABC, given a = 89, b = 57 and C 75°

4' 15'' to find the other parts.