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And since DH is perpendicular to both HB and HE, it is perpendicular to the plain EHB (2. 2 Sup.), therefore the plain DAB which passes by DH is perpendicular to the plain EHB (9.2 Sup. ), and so the plain EHB is perpendicular to the plain DAB, as is manifest from Def. 4. 2 Snp. ; but the plain DBC or DBE is, because of the spherical right angle at B, perpendicular to the same plain DAB (Def. 3. Sph. Tr.); therefore BE the common section of the plains HBE and DBE is perpendicular to the plain DAB (10. 2 Sup.), and EBD and EBH are right angles (Def. 3. 2 Sup.); therefore BE is a tangent of the arch BČ to the same radius (Def. 8 Pl. Tr).

In like manner CG may be shewn to be at right angles to the plain DAB, therefore CGD and CGF are right angles, and CG is the sine of the arch CB to the same radius.

And in the triangle CGF right angled at G, CG is to CF, as the sine of the angle CFG, or (Def. 3 Sph. Tr.), of the spherical angle CAB, is to radius (1. Pl. Tr.); but, as has been just shewn, CG is to CF, as the sine of CB is to the sine of ČA; therefore (11. 5 Eu.), the sine of the angle CAB is to radius, as the sine of CB is to the sine of CA; and therefore the rectangle under radius and the sine of CB, is equal to the rectangle under the sine of CAB and the sine of CA (16. 6 Eu).

PROP. XXII. THEOR.

In a right angled spherical triangle ( ABC, right angled at B, see

Jg. to prec. prop. ), the rectangle under radius and the tangent of either leg (BC), is equal to the rectangle under the tangent of the angle (CAB) opposite that leg, and the sine of the other leg (AB).

The same construction remaining, as in the preceding proposition, in the triangle EBH right angled at B, BE is to BH, as the tangent of the angle EHB, or (Def. 3 Sph Tr.) of the spherical angle CAB, is to radius (2. Pl. Ir.); but, as has been shewn in the preceding prop. BE is to BH, as the tangent of BC is to the sine of BA ; therefore (11. 5 Eu.), the tangent of the angle CAB is to radius, as the tangent of BC is to the sine of BA, and therefore the rectangle under radius and the tangent of Cis,

is equal to the rectangle under the tangent of CAB and the sine of AB (16. 6 Eu).

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If in a right angled spherical triangle, either of the legs, the comple

ment of the hypothenuse, or of either of the angles, except the right angle, be taken as a middle part; the rectangle under radius and the sine of that middle part, is equal to the rectangle under the cosines of those two of the other parts, which, the right angle being excluded, are remote from that middle part.

Let ABC be a right angled spherical

N P triangle, right angled at B. and let either of the legs, AB or BC, the complement of the hypothenuse AC, or of either of the angles except the right angle, name

E ly, of the angle A or ACB, be taken as a middle part; the rectangle under radius and the sine of that middle part, is equal to the rectangle under the cosines of those two of the other parts, which, the A right angle being excluded, are remote

B from that middle part.

Let the great circle DF be described of which A is the pole, meeting AB, AC and BC produced in D, E and F; and since the great circle AD passes through the pole A of the great circle DE, the angle D is a right angle

(Cor. 7. 2 Sph. Tr.), and therefore DE passes through the pole of the arch ABD (Cor. 6. 2 Sph. Tr.); and, because the angle ABC is a right angle, BCF passes also through the pole of ABD (by the same); therefore the arches AD, AE, FB and FD are quadrants (2 Sph. Tr.), and the angle CEF is a right angle (Cor. 7. 2 Sph. Tr.); therefore in the triangle CEF right angled at E, CE is the complement of the hypothenuse AC of the triangle ABC; also FC is the complement of BC, EF of ED, the measure of the angle A (3 Sph. Tr.), and BD, which is the measure of the angle F by the same), is the complement of AB. These things being premised.

Case 1. When either leg, as AB, is the middle part, and the complements of AC and ACB, the remote parts.

, The rectangle under radius and the sine of AB, is equal to the rectangle under the sines of the angle ACB and of the hypothenuse AC (21 Spl. Tr).

Case 2. When the complement of either of the oblique angles, as A, is the middle part, and BC and the complement of ACB, the remote parts.

In the triangle CEF, right angled at E, the rectangle under radius and the sine of EF is, by 21 Sph. Tr., equal to the rectangle under the sine of ECF, or its equal (5. Sph. Tr.), ACB, and the sine of CF; and therefore, substituting the equals in the triangle ABC, the rectangle under radius and the cosine of A, is equal to the rectangle under the sine of the angle ACB and the cosine of BC.

Case 3. When the complement of the hypothenuse AC is the middle part, and the legs AB and BC, the remote parts.

In the triangle CEF, right angled at E, the rectangle under radius and the sine of CE is, by 21. Sph. Tr, equal to the rectangle under the sines of F and FC; and therefore, substituting the equals in the triangle ABC, the rectangle under radius and the cosine of AC, is equal to the rectangle under the cosines of AB and AC.

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The same things being supposed, the rectangle under radius and

the sine of the middle part, is equal to the rectangle under the tangents of the parts, which, the right angle being excluded, are adjacent to the middle part.

The same construction remaining, as in the preceding proposition.

Case 1. When either leg, as AB, is the middle part, and BC and the complement of A, adjacent parts.

The rectangle under radius and the tangent of BC, is equal to the rectangle under the tangent of A and sine of AB (22 Sph. Tr.), therefore the sine of AB is to radius, as the tangent of BC is to the tangent of A (16. 6 Eu.) ; but radius is to the cotangent of A, as the tangent of A is to the radius (Cor. 4 Def. Pl. Tr. and Theor. 3. 15. 5 Eu.); therefore, by equality, the sine of AB is to the cotangent of A, as the tangent of BC is to radius (22.5 Eu.); therefore the rectangle under radius and the sine of AB, is equal to the rectangle under the cotangent of A and the tangent of BC (16. 6 Eu).

Case 2. When the complement of either of the oblique angles, as A, is the middle part, and AB and the complement of AC, adjacent parts.

; and

In the triangle CEF, right angled at E, the rectangle under radius and the tangent of CE, is equal to the rectangle under the tangent of F and the sine of EF (22 ph. Tr.), therefore the sine of EF is to radius, as the tangent of CE is to the tangent of F '16. 6 Eu.); and therefore, substituting the equals in the triangle ABC, the cosine of A is to radius, as the cotangent of AC is to the cotangent of AB, or, tangents being reciprocally as their cotangents (Cor. 5 Def. Pl. Tr.), as the tangent of AB is to the tangent of AC ; and radius is to the cotangent of AC, as the tangent of AC is to radius (Cor. 4 Def. Pl. Tr. and Theor. 3. 15. 5 Eu.); therefore, by equality, the cosine of A is to the cotangent of AC, as the tangent of AB is to radius (22. 5 Eu. therefore the rectangle under radius and the cosine of A, is equal to the rectangle under the cotangent of AC and the tangent of AB (16.6 Eu).

Case 3. When the complement of the hypothenuse AC is the middle part, and the complements of the angles A and ACB, adjacent parts.

In the triangle CEF, right-angled at E, the rectangle under radius and the tangent of EF, is equal to the rectangle under the tangent of ECF and the sine of CE (22. Sph. Tr.), or substituting the equals in the triangle ABC, the rectangle under radius and the cotangent of A, is equal to the rectangle under the tangent of ACR and the cosine of AC; therefore the cosine of AC is to radius, as the cotangent of A is to the tangent of ACB (16. 6 Eu.); and radius is to the cotangent of ACB, as the tangent of ACB is to radius (Cor. 4 Def. Pl. Tr. and Theor. . 3, 15. 5 Eu.); therefore, hy equality, the cosine of AC is to the cotangent of ACB, as the cotangent of A is to radius (22. 5 Eu.), and of course the rectangle under radius and the cosine of AC, is equal to the rectangle under the cotangents of A and ACB (16. 6 Eu.

Scholium.- In any right angled spherical triangle, the five parts mentioned in this proposition and the preceding, namely, the two legs, and the complements of the hypothenuse and oblique angles, are called, circular parts ; of which, any one being considered as a middle part, those, which, the right angle being excluded, are adjacent thereto, are called adjacent parts, and the other two, remote parts, as in these propositions ; both the adjacent and remote parts being called by a common name, extreme purts,

PROP. XXV. THEOR.

If in an oblique anged spherical triangle (ABC), two right angled

spherical triangles (ADC and BDC) be formed, by letting fall a perpendicular (CD), on any side (AB) considered as the base, from the opposite angle, and the perpendicular (CD) be assumed as the middle part in each of the right angled triangles ; the rectangle under the cosines of the remote parts in one of the right angled triangles, is equal to that under the cosines of the remote parts in the other ; and the rectangle under the tangents of the adjacent parts in one, is equal to that under the tangents of the adjacent parts in the other.

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For the rectangles under the cosines of the remote parts in the triangle ADC, which are the complements of A and AC, and in the triangle BDC, which are the complements of CBD and BC, are each of them equal to the rectangle under radius and the sine of the middle part CD (23 Sph. Tr.), and therefore to each other.

And the rectangle under the tangents of the adjacent parts in the triangle ADC, which are AD and the complement of ACD, and in the triangle BDC, which are BD and the complement of BCD, are each of them equal to the rectangle under radius and the sine of the middle part CD (24. Sph. Tr.), and therefore to each other.

Cor. The tangents of the segments of the base [AD and DB are to each other, as the tangents of the vertical angles [ACD and BCD]

For, by this proposition, the rectangle under the tangent of - AD and cotangent of ACD is equal to the rectangle under the tangent of DB and cotangent of BCD; therefore the tangent of AD is to the tangent of DB, as the cotangent of BCD is to the cotangent of ACD (16.6 Ext.), or, the tangents of any two arches

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