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

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52. In any triangle the sides are proportional to the sines of the opposite angles.

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Let ABC be any triangle, and from A draw AD perpendicular to BC or BC produced.

From the right-angled triangles ADB, ADC we have

AD=c sinB,

AD=b sinA CD = b sin C,

since in (1) angle ACD = C, and in (2) ACD=180° – C.

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53. In every triangle any side is equal to the sum of the products obtained by multiplying each of the other sides into the cosine of the angle between it and the first side.

From fig. (1) of Art. 52, we have

a=BD+DC=c cosB+b cos C.

From fig. (2) of Art. 52,

a = BD-DC=c cosB-b cos(180° – C)

Similarly,

=c cosB+b cos C.

b=a cos C+ c cosA,

c=b cosA+ a cosB.

54. To express the cosine of an angle of a triangle in terms of the sides.

See fig. (1) of Art. 52.

Let ABC be a triangle, having the angle Cacute; then (Euclid 11. 13)

c2 = a2 + b2 — 2a. CD;

but CD=b cos C, from the triangle ADC,

.. c2= a2 + b2 - 2ab cos C.

See fig. (2) of Art 52.

Next, let C be obtuse; then (Euclid 11. 12)

c2=a2+b2+2a. CD;

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but CD=b cos(180° C)=-b cos C, from the triangle ADC, .. c2= a2+b2 - 2ab cos C, which is the same as before.

Lastly, let

C=90°; then cos C = cos 90° = 0, and c2=a2+ b2 – 2ab cos C reduces to c2 = a2+b2, which we know to be true (Euclid 1. 47). Therefore in all cases we have

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#55. Given, a=b cos C+c cosB......(1), b=c cosA+ a cos C......(2),

c = a cosB+b cosД...... (3),

a2+b2-c2

to deduce cos C =

2ab

a (1) + b (2) − c (3) gives

and the two similar formulae.

a2+b2-c2= (ab cos C+ ac cosB) + (bc cosA+ ab cos C)

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Similarly the values of cosB and cosA can be deduced.

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sinA =sin(180°-A)=sin(B+ C)=sinB cos C+ cosB sin C.

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The sign

(b2 + c2-a2)2

4b2c2

2a2b2 + 2b2c2 + 2c2a2 — aa — ba1 — c1

462c2

+√(2a2b3 +2b3c2 +2c3a2 — a* — b* — c1)

2abc

is placed before the radical sign because A is

less than 180°, and therefore sin A is positive.

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58. To express the sine, cosine, and tangent of half an angle of a triangle in terms of the sides.

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(Art. 37);

Now 1-cosA = 2 sin2, and 1+ cos▲ = 2 cos31⁄2▲,

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since

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s(s-a)

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The positive sign must be given to all the above radicals, is less than 90°, and therefore all its Trigonometrical Ratios are positive.

COR. Hence sinA = 2 cos4 sin

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To investigate the three following formulae:

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