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=

.. the angle ACD = { of a right-
angle = 30°,
and the angle DAC= of 2 right-
angles = 60°.
Also, as in Euc. I. 10, the tri-

2
angles DAC, DBC are equal in
all respects.
:. AD = DB; and DC is at right-

A 1 D

B angles to AB.

Take AD for unit length.
Then

AC = AB= 2. AD = 2.
And
DC2 + AD2 = AC2.

(Euc. I. 47.
i.e. DC2 + 1 = 22 = 4,
.. DC2 = 3 and DC = (3.
DC 73

AD 1
Thus, sin 60°
AC 2

AC 2
AD 1

DC J3

13 and sin 30° AC2

AC

2 Since 30° and 60° are complementary, any ratio of the one is equal to the co-ratio of the other. 109. The ratios of 1 and of a right-angle.

пс 37

and 8 8 .

: cos 60°

: &c.,

: cos 30°

: &c.

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Describe the isosceles triangle ABC, right-angled at A. Produce AB to D, making BD = BC. Join DC.

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2210

45°

Then, as in Art. 107, _ ABC is a right-angle; and the angles BDC, BCD being equal by Euc. I. 5, each of them is of a right-angle by Euc. I. 32. Take A B = AC=1. Then BD = BC = 72, as in Art. 107.

DA
.. cot 223° = cot ADC = = 12 +1.

AC
Again DC? - DA+ AC"? = (2+2 /2 + 1) + 1 = 4 + 2 /2.

DC
.. cosec 221° = cosec ADC = (4+2 J2).

AC Similarly the other ratios may be written down, and their denominators may be rationalised by an algebraic process.

The same figure gives the ratios of 671

By the following simple geometrical construction; the other ratios may be at once written down with rationalised denominators.

With centre B and radius BD or BC, describe the semi-circle DCE, cutting BA produced in E. Join CE. .

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..

Then _ DCE in a semi-circle is a right-angle.

:: LACE=complement of LDCA= L ADC=22 Also

AE=BE - BA=2-1, .:: CE2= A E2+ AC2=(2–2./2+1)+1=4-2,12, :: from A EDC, sin 22° = 1/(2-2) and cos 22°=1/(2+J2), and from a ECA, tan 22° = 2-1 and sec 22° = (4-2/2).

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Let ABC be the half of an equilateral triangle, so that, as in Art. 108, BCA = 60°, L ABC = 30°, BAC = 90°. Produce AB to D, making BD= BC. Join DC. Then, the angles BDC, BCD being equal,

[Euc. I. 5. each of them is half angle ABC, i.e. 15°.

[Euc. I. 32. Take AC=1. Then BD=BC = 2, and AB= 73, as in Art. 108.

DA .. cot 15o = cot ADC

2 + 13. AC

=

=

=

Again, DCP = DA' + AC? =(4+ 4/3 + 3) + 1 = 8 + 4/3 = 6 + 2/12 + 2,

:. DC= /6+ /2.

DC
: . cosec 15°
cosec ADC = = 16+ /2.

AC

Similarly the other ratios of 15° and 75° may be written down, and their denominators may be rationalised by an algebraic process.

By the same construction as in the preceding article, the other ratios may be at once written down with rationalised denominators.

With centre B and radius BD or BC, describe the semi-circle DCE, cutting BA produced in E. Join CE. Then _ DCE in a semi-circle is a right-angle,

LACE=complement of LDCA= LADC=15°.

..

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Also

AE=BE-BA=2-/3, .:: CE2=A E2+ AC2=(4-4/3+3)+1=8-42/3, .: from A EDC, sin 15°=I(26-2) and cos 15o =$(/6+2),

and from a ECA, tan 15°=2-13 and sec 15°= 16-12.

+

111. We may proceed to bisect the angles 222° and 15° in the same way.

The student will easily show that
cot 117° = 12 +1 + 7 (4+2/2),

cot 74° = 2+ 3+ /6+ /2. In precisely the same way the student may prove geometrically the general formula

cot A = cosec 2A + cot 2A.

2

3

112. The ratios of i, j, 3, of a right angle.

[18°, 36°, 54°, 72°.] Bisect any straight line AB in C. Draw BD at right angles and equal to AB. Join CD. Produce AB to E, making CE equal to CD. With centres A and B and radii each equal to AE describe two circles cutting in F. Join FA, FB, so that FA and FB are each equal to AE. The angle A FB shall be 36o. For :: AB is bisected in C and produced to E, :. AE. BE + BC = CE2 = CD2 = BD + BC?,

.. AE. BE=BD2 or ABP.

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From FB, which is equal to A E, cut off FO equal to AB; then the remainder BO = remainder BE. :. the above equation becomes BF. BO= AB,

:. BF: AB=AB : BO, .. the triangles FBA, ABO, having the sides about their common angle B proportional, are similar.

.. LAOB= FAB - FBA and AO=AB=FO. But :: A0= FO, ..ZAFO = 2FAO, ...AOB = twice - AFO. J. T.

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