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EXAMPLES.

1. Calculate the value of the tangent of 60°.

sin 60°

V3 tan 60°

• j = V3.

Ans. tan 60° = V3 = 1.73205. 2. Calculate the value of the secant of 30°.

cos 60°

2

I

V等

sec 30°

cos 30°

Ans sec 30o = 1.15470. 3. Calculate the value of the tangent of 30°.

Ans, tan 30o = 0.57735. 4. Calculate the sine and cosine of 120°. As 120° is the supplement of 60°, it follows from sect. 3, that

Ans. sin i 20° sin 60° = 0.86602,
COS I 20o =- cos 60° =-

- 0.50000. 5. Calculate the tangent of 120°.

sin i 20° sin 60° tan 120°

tan 60°.

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COS I 20°

cos 60°

-I

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2

Ans. tan 120° 1.73205. 6. Calculate the sine of 18°.

If ACB and ACB' (fig. 7) be each 18°, BCB' is equal 36°, and as 36° is contained ten times in 360°, it follows that BB' is the side of a regular decagon inscribed in the circle, and therefore that it is equal to the greater segment of the radius (= 1), cut in extreme and mean ratio (Euclid, Book IV. Prop. x.); but the greater part

V5of unity cut in extreme and mean ratio

and as BP, or the sine of 18°, is half BB, it follows that

V5 -
Ans. sin 18° =

4

= 0.30901. 7. Calculate the cosine of 18o.

V(10 + 2V5) Ans. cos 18° =

4

= 0.95105. 8. Calculate the tangent of 18o.

V5 - 1
Ans. tan 18° =

V(10 + 2V5) = 0.32492.

9. Calculate the sine, cosine, and tangent of 72°. Since 72° is the complement of 18°,

Ans. sin 72° = 0.95195,

cos 72° = 0.30901,
tan 72° = 3.07768

CHAPTER III.

RIGHT-ANGLED TRIANGLES.

1. Relations between the sides and Angles.—2. Four Cases of

Right-Angled Triangles.-3. The Four Cases computed by
Logarithmic Tables.

1. Relations between the sides and angles. -Let ABC (fig. 8) be a right-angled triangle, and let AN be measured off upon the hypotenuse, equal in length to the linear unit. With A as centre, and AN as radius, describe an arc of a circle, NM; draw NP and MT perpendicular to AC, then NP is the sine and MT the tangent of the angle A. Since the triangles BÃC and NAP are similar,

BC : AB :: NP: AN,

:: sin A :1; and therefore

BC = AB x sin A. As the angle B is the complement of A, it follows (Chap. II. sect. 3) that

BC = AB x cos B. Similar values may be obtained for the side AC, * and therefore

PROPOSITION I. In a right-angled triangle, either side is equal to the hypotenuse, multiplied by the sine of the opposite, or cosine of adjacent, angle.

:

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Since the triangles BAC and TAM are similar,
BC : AC :: TM : AM,

:: tan A:1; and therefore

BC = AC x tan A. As A and B are complemental, it follows (Chap. II. sect. 3) that

BC = AC x cot B. Similar values may be obtained for the side AC, and therefore,

PROPOSITION II.

a

In a right-angled triangle, either side is equal to the other multiplied by the tangent of the opposite, or cotangent of adjacent, angle.

In a right-angled triangle there are five quantities, viz.: the two sides, the hypotenuse, and the two angles adjacent to the hypotenuse. Let these be denoted by the letters a, b, c, A, B, as in (fig. 9):

The propositions established in the last section may be expressed by the equation: a = c sin A = c cos B,

(1) 6 -- c sin B = c cos A;

(2) and b tan A = b cot B,

(3) b = a tan B = a cot A.

(4) 2. Four cases of right-angled triangles.If any two of the five quantities, a, b, c, A, and B, be given, the remaining three may be calculated Four cases may be proposed, viz. :

a =

CASE I. Given the two sides.
CASE II.—Given one side and hypotenuse.
CASE III.—Given one side and either angle.
CASE IV.–Given hypotenuse and either angle.

The following examples may be solved by means of the Tables I., II., III., in the Appendix :

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CASE I.

Given a and b, it is required to find c, A, and B. By equation (3),

a

tan A = 0;

tan A having been calculated from this equation, A may be found from the tables,

B = 90° – A; and c may be calculated from the equation,

C= v(a? + 62). -(Euclid, Book I. Prop. XLVII.)

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EXAMPLES.

1. Given the two sides of a right-angled triangle, equal to 500 feet and 70.27 feet respectively, find the base angles and the hypotenuse. Let a= 70.27 and b = 500; then tan A=0.14054, which corresponds in the tables to 8°; the complement of this 82° = B. If the squares of a and b be added, the sum is 254937.8729; the square root of this being taken, we find 504.91.

Ans. c = 504.91 feet.

A= 8°. B= 82°. 2. Given a = 141.407, and b = 350; find c, A, and B.

C,
Ans. c=

377.486.

A = 22°. B = 68°. 3. Given a = 127.38, b = 250; find c, 4, and B.

Ans. C=

= 280.58. 4 = 27°. B = 63°.

CASE II.

Given a and c, it is required to find b, A, and B. From equation (1),

sin A

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Sin A having been calculated from this equation, A may be found from the tables,

B = 90° - A; and b may be calculated from the equation,

b = V(c– a). --(Euclid, Book I. Prop. xLvII.)

EXAMPLES.

1. Given a = 5.1303, C= 15; find b, A, and B.

Ans. b= 14.09539.

A = 20°. B= 70°. 2. Given a = 128.76, c= 250 ; find b, A, and B.

Ans. b = 214.29.

4 = 31°. B = 59°. 3. Given a = 141.57675, c = 175; find b, A, and B.

Ans. b = 102.86.

A = 54°. B = 36°.

CASE III.

Given a and A, it is required to find B, b, and c. By equation (4),

b =

= a cot A. By equation (1),

a

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