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1. Calculate the value of the tangent of 60°.
V3 tan 60°
• j = V3.
Ans. tan 60° = V3 = 1.73205. 2. Calculate the value of the secant of 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,
- 0.50000. 5. Calculate the tangent of 120°.
sin i 20° sin 60° tan 120°
COS I 20°
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
= 0.30901. 7. Calculate the cosine of 18o.
V(10 + 2V5) Ans. cos 18° =
= 0.95105. 8. Calculate the tangent of 18o.
V5 - 1
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,
1. Relations between the sides and Angles.—2. Four Cases of
Right-Angled Triangles.-3. The Four Cases computed by
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.
Since the triangles BAC and TAM are similar,
:: 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,
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. :
CASE I. Given the two sides.
The following examples may be solved by means of the Tables I., II., III., in the Appendix :
Given a and b, it is required to find c, A, and B. By equation (3),
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.)
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.
A = 22°. B = 68°. 3. Given a = 127.38, b = 250; find c, 4, and B.
= 280.58. 4 = 27°. B = 63°.
Given a and c, it is required to find b, A, and B. From equation (1),
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.)
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°.
Given a and A, it is required to find B, b, and c. By equation (4),
= a cot A. By equation (1),