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In a right angled triangle, the radius is to the tangent of an oblique angle, as the adjacent side to the opposite side.

Let the triangle ABC be right angled at B; then R : tamb AC : : AB : BC.

For, assuming AR equal to the given radius, describe the arc RD, and draw the perpendicular RT. The triangles C ART and ABC being similar, AR : RT: ; AB : BC. But, AR being the radius, RT is the tangent of the arc RD which measures the angle at A ; and therefore R tanA : : Ā R. B. AB : BC,

Cor. Hence the radius is to the secant of an angle, as the adjacent side to the hypotenuse. For AT is the secant of the arc RD, or of the angle at A ; and, from similar triangles, AR : AT: ; AB : AC.


The sides of any triangle are as the sines of their opposite angles.

In the triangle ABC, the side AB is to BC, as the sine of the angle at C to the sine of that at A.


For let a circle be described about the triangle; and the sides AB and BC, being chords - of the intercepted arcs or of the angles at the centre, are(cor. def.) equal to twice the sines of the halves of those angles, or the angles ACB and CAB at the circumference. But, of the same angles, the chords or sines (VI. 11. cor. El.) are proportional to the "radius; and consequently AB : BC : : sinc : sin.A.

Cor. Since the straight lines AB and BC are chords, not only of the arcs AB and BC, but of the arcs ACB and BAC, or the defects of the former from the circumference, it follows that the sides of the triangle are proportional also to the sines of half these compound arcs, or to the sines of the supplements of their opposite angles.—A like inference results from the definition, for the sine of an arc and that of its supplement are the same.


In any triangle, the sum of two sides, is to the difference, as the tangent of half the sum of the angles at the base, to the tangent of half their dif. ference.

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From the vertex A, and with a distance equal to the greater side AB, describe the semicircle FBD, meeting the other side AC extended both ways to F and D, join BD and BF, which produce to meet a straight line DE drawn

parallel to CB.

Because the isosceles triangle DAB, has the same vertical angle with the triangle CAB, each of its remaining angles ADB and ABD is (I. 30. El.) equal to half the sum of the angles ACB and ABC; and therefore the defect of ABC from that mean, that is the angle CBD, or its alternate angle BDE, must be equal to half the difference of those angles. Now FBD being (III. 19. El.) a right angle, BF and BE are tangents of the angles BDF and BDE, to the radius DB, and hence are proportional to the tangents of those angles with any other, radius. But since CB and DE are parallel, CF, or AB+AC : CD, or AB–AC :: BF : BE; consequently

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Cor. Suppose another triangle abc to have the sides ab and ac equal to AB and AC, but containing a b

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In any triangle, as twice the rectangle under two sides, is to the difference between their squares and the square of the base, so is the ra. dius to the cosine of the contained angle.

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In any triangle, the rectangle under the semiperimeter and its excess above the base, is to the rectangle under its excesses above the two sides, as the square of the radius, to the square of the tangent of half the contained angle.

In the triangle ABC, the perimeter being denoted by P, 3|P(AP–AC : (AP—AB) (; P-BC): ; Ro : tank B2.

For, employing the construction of Prop. 29., Book VI. of the Elements; since the triangles BIE and BGD

are right angled, BI: IE :

and BG : G D whence (V. 22. El.) BI.BG But it was proved that IE.GD=AI.AG; wherefore BI.BG : AI.A.G. : : R* : tank B". Now BI is equal to the semiperimeter, BG is its excess above the base AC, and AI, AG are its excesses above the sides AB and BC; consequently the proportion is



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