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PROP. XXIII.
In right angled spherical triangles, the cosine of either of the sides is to

the radius, as the cosine of the angle opposite to that side is to the
sine of the other angle.

COS

The same construction remaining : In the triangle CEF, sin CF : R :: sin EF : sin ECF, (19.); but sin CF

= cos CA, sin EF ABC, and sin ECF=sin BCA; therefore, cos CA :R :: cos ABC : sin BCA. Q. E. D.

PROP. XXIV.

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In spherical triangles, whether right ungled or oblique angled, the sinės of the sides are proportional to the sines of the angles opposite to them.

First, Let ABC be a right angled triangle, having a right angle at A; therefore, (19.) the sine of the hypotenuse BC is to the radius, (or the sine of the right angle at A),

С as the sine of the side AC to the sine of the angle B. And, in like manner, the sine of BC is to the sine of the angle A, as the sine of

B AB to the sine of the angle C; wherefore (11.5.) the sine of the side AC is to the sine of the angle

A B, as the sine of AB to the sine of the angle C.

Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the sine of the angle A opposite to BC, is to the sine of the angle B opposite to AC. Through the point C, let there be drawn an arch of a great circle CD perpendicular to AB; and in the right angled trian

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D
B D

A
gle BCD, sin BC :R :: sin CD : sin B, (19.); and in the triangle
ADC, sin AC :R :: sin CD:sin A ; wherefore, by equality inversely,
sin BC : sin AC :: sin A : sin B. In the same manner, it may be
proved that sin BC': sin AB :: sin A : sip C; &c. Therefore, &c.
Q. E. D.

PROP. XXV.'

In oblique angled spherical triangles, a perpendicular arch being drawn

from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle.

Let ABC be a triangle, and the arch CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD.

For having drawn CD perpendicular to AB, in the right angled triangle BCD, (23.) cos CD : R :: cos B : sin DCB ; and in the right angled triangle ACD, cos CD : R :: cos A : sin ACD; therefore (11.5.) cos B : sin DCB :: cos A : sin ACD, and alternately, cos B : cos A :: sin BCD : sin ACD. Q. E. D.

PROP. XXVI.

The same things remaining, the

cosines of the sides BC, CA, are proportional to the cosines of BD, DA, the segments of the base.

For in the triangle BCD, (22.), cos BC : cos BD :: cos DC : R, and in the triangle ACD, cos AC :cos AD :: cos DC : R; therefore (11. 5.) cos BC : cos BD :: cos AC : cos AD, and alternately, cos BC : cos AC :: cos BD : cos AD. Q. E, D.

PROP. XXVII.

The same construction remaining, the sines of BD, DA the segments of the

base are reciprocally proportional to the tangents of B and A, the angles at the base.

In the triangle BCD, (18.), sin BD: R :: tan DC : tan B; and in the triangle ACD, sin AD: R : : tan DC : tan A; therefore, by equality inversely, sin BD : sin AD : : tan A : tan B. R. E. D.

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PROP. XXVIII.

The same construction remaining, the cosines of the segments of the verti

cal angle are reciprocally proportional to the tangents of the sides.

Because (21.), cos BCD : R :: tan CD : tan BC, and also, cos ACD : R :: tan CD : tan AC, by equality inversely, cos BCD : cos ACD : : tan AC : tan BC. Q. E. D.

PROP. XXIX.

COS M-COS n.

If from an angle of a spherical triangle there be drawn a perpendicu

lar to the opposite side, or base, the rectangle contained by the tangents of half the sum, and of half the difference of the segments of the base is equal to the rectangle contained by the tangents of half the sum, and of half the difference of the two sides of the triangle.

Let ABC be a spherical triangle, and let the arch CD be drawn from the angle C at right angles to the base AB, tan į (m+n) X tan į (m-n) = 1 tan (a+b)x tan (a-b).

Let BC=a, AC=b; DD=m, AD=n. Because (26.) cos a : cos b:: cos m : cos n, (F. 5.) cos a +b: cos a-cos b :: cos m+cos n :

But' (1. Cor. 3. Pl. Trig.), cos a +cos b : cos a -- cos b :: cot} (a+b): tan į (a−b), and also, cos m+cos n : cos m-cos n :: cotį (m+ii) : tan (m-n). Therefore, (11. 5.) cot } (a+b): tan (a-6):: cot (m+n) : tan $ (m-n). And because rectangles of the same altitude are as their bases, tan ļ (a + b) X cot } (a+b): tan (a+b) X tan} (ü -b) :: tan} (m + n) Xcot (m+n) : tan (m Xn) + tan < (m,n). Now the first and third terms of this proportion are equal, being each equal to the square of the radius, (1. Cor. Pl. Trig.), therefore the remaining two are equal, (9. 5.) or tan (m + n) x tan į (m-n) =tan } (a + b) Xtan} (a-b), that is, tan (BD + AD) X tan (BD-AD) = tan; (BC + AC) X tan} (BC-AC). Q. É. D.

Cor. 1. Because the sides of equal rectangles are reciprocally proportional, tan }(BD + AD) : tan} (BC+AC) :: tan; (BC-AC): tan (BD-AD).

Cor. 2. Since, when the perpendicular CD falls within the trian. gle, BD + AD = AB, the base ; and when CD falls without the triangle BD-AD = AB, therefore in the first case, the proportion in the last corollary becomes, tan Į (AB): tan } (BC + AC) :: tan į (BC-AC): tan (BD-AD); and in the second case, it becomes by inversion and alternation, tan · (AB) : tan ļ (BC + AC) :: tan į (BC-AC): tan(BD+AD).

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The preceding proposition, which is very useful in spherical trigo nometry, may be easily remembered from its analogy to the proposition in plane trigonometry, that the rectangle under half the sum, and half the difference of the sides of a plane triangle, is equal to the rectangle under half the sum, and half the difference of the seg. ments of the base. See (K. 6.), also 4th Case Pl. Tr. We are indebted to NAPIER for this and the two following theorems, which are so well adapted to calculation by Logarithms,athat they must be con

as three of the most valuable propositions in Trigonometry:

PROP. XXX.

If a perpendicular be drawn from an angle of a spherical triangle to the

opposite side or base, the sine of the sum of the angles at the base is to the sine of their difference as the tangent of half the base to the tangent of half the difference of its segments, when the perpendicular fulls within; but as the co-tangent of half the base to the co-tangent of half the sum of the segments, when the perpendicular falls without the triangle: And the sine of the sum of the two sides is to the sine of their difference as the co-tangent of half the angle contained by the sides, to the tangent of half the difference of the angles which the perpendicular makes with the same sides, when it falls within, or to the tangent of half the sum of these angles, when it falls without the triangle.

If ABC be a spherical triangle, and AD a perpendicular to the base BC, sịn (C+B): sin (C-B) :: tan į BC : tan < (BD-DC), when AD falls within the triangle ; but sin (C+B) : sin (C-B) :: cot BC : cot } (BD+DC), when AD falls without. And again,

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D

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D
B

B В sin (AB+AC): sin (AB-AC) :: cot | BAC : tan} (BAD-CAD), when AD falls within ; but when AD falls without the triangle, $in (AB+AC) : sin (AB-AC :: cot? BAC : tan (BAD+CAD).

For in the triangle BAC (27.), tan B : tan C :: sin CD : sin BD, and therefore (E. 5.), tan C + tan B : tan C-tan B :: sin BP + sin CD : sin BD-sin có. Now, (by the annexed Lemma) tan c + tan B : tan C-tan B :: sin (C+B): sin (C-B), and sin BD+ sin CD: sin BD-sin CD :: tan } BD+CD): tan Į (BD-CD),(3. Pl. Trig.), therefore, because ratios which are equal to the same ratio are equal to one another (11.5.), sin (C+B): sin (C-B) :: tanį (BD+CD): tan I (BD-CD).

A

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B

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B .Now when AD is within the triangle, BD+CD=BC, and therefore sin (C+B): sin (C-B) :: tan1 BC : tan (BD-CD). And again, when AD is without the triangle, BD-CD=BC, and therefore sin (C+B) : sin (C-B) :: tan (BD+CD): tan BC, or because the tangents of any two arches are reciprocally as their co-tangents, sin (C+B) : sin (C-B) :: cot } BC : cot } (BD+CD).

The second part of the proposition is next to be demonstrated. Because (28.) tan AB : tan AC :: cos CAD : cos BAD, tan AB + tan AC : tan AB-tan AC :: cos CAD+cos BAD : cos CAD-cos BAD. But (Lemma) tan AB + tan AC : tan AB-tan AC :: sin (AB+AC): sin (ÀB- AC), and (1. cor. 3. Pl. Trig.) cos CAD + cos BAD : cos CAD-cosBAD :: cot(BAD+CAD) : tan(BAD-CAD). Therefore (11. 5.) sin (AB+AC): sin (AB-AC) :: cot} (BAD+CAD): tan (BAD-CAD). Now, when AD is within the triangle, BAD-CAD=BAC, and therefore sin (AB+-AC): sin (AB-AC) :: cot : BAC : tan (BAD-CAD).

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