Fig. 317.

Having, in this case, the base and perpendicular known, by referring to the above proportions we find that with these two sides we may obtain the tangent; therefore -

Tan. B = b/a = 6/8 =0-75.

Referring to the trigonometrical tables, we find that 0.75 is the tangent of 360 52' 12", nearly; therefore -

 The quadrant equals 90.0.0 The angle B equals 36.52.12 The angle A equals 53.07.48

475. - Right-Angled Triangles: Trigonometrical Value of Sides. - In the triangle A BC (Fig. 317), with B P = 1 for radius, and on B as a centre, describe the arc PD, and from its intersection with the lines A B and B C, draw PM and TD perpendicular to the line B C. Then from homologous triangles we have these proportions for the perpendicular -

Bd: D T:: Bc: Ca,

r: tan. B:: base: perp.,

1 : tan. B:: a: b - a tan. B. (185.)

Also -

Bp: Pm:: Ba: A C,

r: sin. B:: hyp.: perp.,

1: sin. B:: c: b = c sin. B. (186.)

For the base, we have -

Bp: Bm:: Ba: B C,

r: cos. B:: hyp.: base,

1: cos. B:: c: a = c cos. B. (187.)

Again

T D: B D:: A C: B C,

tan. B: r:: perp.: base,

tan. B: 1:: b: a =b/tan.B. (188.)

For the hypothenuse, we have -

PM: PB:: AC:: AB, sin. B: r:: perp.: hyp.,

sin. B: 1 :: b: c = b/sin.B. (189.)

Again -

Bd: B T:: Bc: Ba,

r: sec. B:: base: hyp.,

I: sec. B:: a: c = a sec. B =a/cos.B . (190.)

This substitution of the cos. for the sec. is needed because tables of secants are not always accessible. That it is an equivalent is clear; for we have -

Bm: Bp:: Bd: B T,

cos.: r:: r: sec. =1/cos.

By these equations either side of a right-angled triangle may be computed, provided there are certain parts of the triangle given. As, for example: of the six parts of a triangle (the three sides and the three angles), three must be given, and at least one of these must be a side.

As an example: let it be required to find two sides of a right-angled triangle of which the base is 100 feet, and the acute angle at the base is 35 degrees. Here we have given one side and two angles (the base, acute angle, and the right angle) to find the other two sides, the perpendicular and the hypothenuse.

Among the above rules we have, in equation (185.), for the perpendicular -

b = a tan. B.

Or: The perpendicular equals the product of the base into the tangent of the acute angle at the base.

Then (Art. 427) -

 The logarithmic tangent of B (= 35o) is 9.84523 Log. of a (= 100) is 2 Perpendicular, b (= 70.02075) = 1.84523

And for the hypothenuse, taking equation (190.), we have -

c= a/cos. B

Or: The hypothenuse equals the quotient of the base divided by the cosine of the acute angle at the base. For this we have -

 Log. of a (= 100) is 2 " cos.B (= 35o) is 9.91336 Hypothenuse c(= 122.0775) = 2.08664

We thus find that a right-angled triangle, having an angle of 35 degrees at the base, has its three sides, the perpendicular, baseband hypothenuse, respectively equal to 70.02075, 100, and 122.0775.

Fig. 318.

N.B. - The angle at A (Fig.517) is obtained by deducting the angle at B from 900 (Art. 346). Thus, 90 - 35 = 55; this is the angle at A, in the above case.

If the perpendicular be given, then for the base use equation (188.), and for the hypothenuse use equation (189.). If the hypothenuse be given, then for the base use equation (187.) and for the perpendicular use equation (186.).

476. - Oblique-Angled Triangles: Sines and Sides. - In the oblique-angled triangle A B C (Fig. 318) from C and perpendicular to AB draw CD. This line divides the oblique-angled triangle into two right-angled triangles, the lines and angles of which may be treated by the rules already given; but there is a still more simple method, as will now be shown.

As shown in Art. 474: "When the perpendicular is divided by the hypothenuse the quotient equals the sine." Applying this to Fig. 318, we have -

sin. A = d/b;

sin. B =d/a.

Let the former be divided by the latter; then -

d

sin. A = b sin. B d ' a

or, reducing, we have -

sin. A = a; sin. B b

or, putting the equation in the form of a proportion-sin. B: sin. A:: b: a;

or; the sines are in proportion as the sides, respectively opposite. Or, as commonly stated, the sines are in proportion as the sides which subtend them.

This is a rule of great utility; by it we obtain the following:

Referring to Fig. 318, we have-sin. B: sin. A :: b: a = b sin.A/sin. B. . (191)

sin. C: sin. A:: c: a = c sin. A/sin. C. (192.)

sin. A: sin. B:: a: b = a sin. B/sin. A . (193.)

sin. C: sin. B:: c: b = c sin. B/sin. C (194.)

sin. A: sin. C:: a: c = a sin. C/sin. A. (195.)

sin. B: sin. C:: b: c = b sin. C/sin. B. (196.)

These expressions give the values of the three sides respectively; two expressions for each, one for each of the two remaining sides; that is to be used which contains the given side.

From these expressions we derive the values of the sines: thus -

sin. A = sin. B a/b. (107.)

sin. A = sin. C a/c. (198.)

sin. B = sin. A b/a. (199.)

sin. B = sin. C b/c. (200.)

sin. C = sin. A c/a. (201.)

sin. C = sin. B c/b. (202.)

477. - Oblique - Angled Triangles: First Class. - The problems arising in the treatment of oblique-angled triangles have been divided into four classes, one of which, the first, will here be referred to. The problems of the first class are those in which a side and two angles are given, to find the remaining angle and sides.

As to the required angle, since the three angles of every triangle amount to just two right angles (Art. 345), or 180o, the third angle may be found simply by deducting the sum of the two given angles from 180o.

For example: referring to Fig. 318, if angle A - 18o and angle B = 42o, then their sum is 18 + 42 = 60, and 180 - 60 = 120o = the angle AC B.