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A

B
D

с (2) Write down the following ratios in the above figure ; (i) sin BAD, (ii) cos ACD, (iii) tan DAC, (iv) sin ABD, (v) tan BAD, (vi) sin DAC, (vii) cos DCA, (viii) tan DCA, (ix) cos ABD, (x) sin ACD.

(3) Let ACB be any angle and let ABC and BDC be right angles; (see next figure). Write down two values for each of the following ratios; (i) sin ACB, (ii) cos ACB, (iii) tan ACB, (iv) sin BAC, (v) cos BAC, (vi) tan BAC.

B

D

(4) In the accompanying figure BDC, CBA and EAC are right angles. Write down (i) sin DBA, (ii) sin BEA, (iii) sin CBD, (iv) cos BAE, (v) cos BAD, (vi) cos CBD, (vii) tan BCD, (viii) tan DBA, (ix) tan BEA, (x) tan CBD, (xi) sin DAB, (xii) sin BA E. (5) Let ABC be a right-angled triangle such that AB=5 ft., BC=3ft., then AC will be 4 ft.

B

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a

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Find the sine, cosine and tangent of the angles at A and B respectively.

In the above triangle if A stand for the angle at A and B for the angle at B, show that sin? A + cos2 A=1, and that sin? B + cos2 B=1.

(6) If ABC be any right-angled triangle with a right angle at C, and let A, B, and C stand for the angles at A, B and C respectively, and let a, b and c be the measures of the sides opposite the angles A, B and C respectively.

b Show that sin A

COS A

tan A=

ī Show also that sin? A + cos2 A=1.

Show also that (i) a=c.sin A, (ii) b=c.sin B, (iii) a=c.cos B, (iv) b=c.cos A, (v) sin A=cos B, (vi) cos A=sin B, (vii) tan A =cot B.

(7) The sides of a right-angled triangle are in the ratio 5 : 12 : 13. Find the sine, cosine and tangent of each acute angle of the triangle.

(8) The sides of a right-angled triangle are in the ratio 1:2:13. Find the sine, cosine and tangent of each acute angle of the triangle.

(9) Prove that if A be either of the angles of the above two triangles sino A + cos2 A=1.

CHAPTER VI.

ON THE TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES.

87. The Trigonometrical Ratios of an angle are numerical quantities simply, as their name ratio implies. They are in nearly all cases incommensurable numbers.

Their practical value has been found for all angles between 0 and 90°, which differ by l'; and a list of these values will be found in any volume of Mathematical Tables.

The student is recommended to get a copy of Chambers' Mathematical Tables for instruction and reference.

88. The finding the values of these Ratios has involved a large amount of labour; but, as the results have been published in Tables, the finding the Trigonometrical Ratios does not form any part of a student's work, except to exemplify the method employed.

89. The general method of finding Trigonometrical Ratios belongs to a more advanced part of the subject than the present, but there are certain angles whose Ratios can be found in a simple manner.

90. To find the sine, cosine and tangent of an angle of 45°.

If one angle of a right-angled triangle be 45', that is, the half of a right angle, the third angle must also be 45'. Hence 45° is one angle of an isosceles right-angled triangle.

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m

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Let POM be an isosceles triangle such that PMO is a right angle, and OM = MP. Then POM = OPM = 45°.

Let the measures of OM and of MP each be m. Let the measure of OP be x. Then

x = m + mo = 2m> ;

12. m.
MP

1 Hence, sin 45o = sin POM=

ОР

12.m 12

OM cos 45o = cos POM =

OP

72.m MP

1 tan 45o = tan POM=

1.
OM

1
91. T'o find the sine, cosine and tangent of 60'.

In an equilateral triangle, each of the equal angles is 60°, because they are each one third of 180°. And if we draw a perpendicular from one of the angular points of the triangle to the opposite side, we get a right-angled triangle in which one angle is 60°.

Let OPQ be an equilateral triangle. Draw PM perpendicular to OQ. Then OQ is bisected in M.

Let the measure of OM be m; then that of OQ is 2m and therefore that of OP is 2m.

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M
Let the measure of MP be x.
Then x* = (2m) - = 4m' - m' = 3m",
.. x = 13. m.

MP V3.m
Hence, sin 60o = sin POM

OP 2m

OM cos 60o = cos POJ=

ОР

2m

MP V3.m tan 60° = tan POM :

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92. To find the sine, cosine and tangent of 30°.

With the same figure and construction as above, we have the angle OPM=30°, since it is a half of OPQ, i.e. of 60°.

MO

1 Hence, sin 30o = sin OPM =

ΡΟ 2m

PM 13.m
cos 30o = cos OPM =

PO 2m 2
MO

1
tan 30o = tan OPM=

PM

=

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13

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