A

=

T's (a+b+c) (a + c − b) (b + a −c) (b + c − a)

= s (s - b) (sc) (sa);

or ▲ = √s (sa) (sb) (s - c) .... (2),

·

the expression for the area in terms of the sides.

Cor. 1. If P1, P2, P3, be the perpendiculars on the sides a, b, c, respectively, from the opposite angles, then

The last is deduced from (1) and (2), and the other two are got in a similar way.

Radius of the Inscribed Circle.-Let O be the centre of the circle inscribed in the triangle A B C, and F, D, E the points of contact with the sides A B, B C, C A. Then because the whole triangle ABC is made up of the triangles A O B, B O C, and A O C, we have

(A B+ B C + AC) OF 2 ABC,

=

=

or (a+b+c) r 2 A; hence
√ 8 (sa) (sb) (sc)

7' =

2 A
a+b+c

the required radius.

S

... (3),

If r, 7, 7 be the radii of the circles which touch, respectively, the sides a, b, c, exteriorly, then

/s (s

The investigation of these, which is exactly similar to that for r, is left for the student's exercise.

Radius of the Circumscribed Circle.-If p, be the perpendicular from C on A B, then (Euc. vi. C.),

PROBLEM VIII.

Given the hypothenuse of a right-angled triangle, and the side of the inscribed square, to find the base and perpendicular, when one of the angles of the square is coincident with the right angle of the triangle.

Let A B C be the right-angled triangle and EDBF the inscribed square, so that two of its sides lie on the base and perpendicular of the triangle. Bisect the given hypothenuse A C in O, and put A E + EC = 2 m, AE EC 2 x, the side of the a; then A E = m +x, EC = m − x

=

A C square E D

and O E

=

= x.

Hence by similar triangles

A

AE: EFA C: C B, or m + x: a :: 2 m: CB =

and EC ED:: AC: AB, or m − x: a : : 2 m : A B =

2 ma

m + x'

2 ma

m-x

(m + x)2 (m − x)2 = (m3 — x2)3.

x − 2 (m2 + a2) x2 = m2 (2 a2 - m2); hence

x =

± {(m2 + a2) ± a √ (4 m2 + a2)}*.

Let us now ascertain how many of these values of x fulfil the conditions of the problem geometrically; and first in respect of the double. sign preceding the radical expression.

The expressions for C B and B A show at once that the triangle is the same (except that A B and B C interchange places), whether the plus or minus sign is used. Thus if the value of x be denoted by a, then

when the negative sign is used.

Again, since m2 + a2 > m2, the negative sign within the radical must be taken, otherwise x will be greater than m, and consequently A B will be negative. The only value of x, then, which fulfils the conditions of the problem geometrically, is

x=

{(m2 + a2) − a √ (4 m2 + a°) }*.

(1).

Scholium. The value of x, when the plus sign within the brackets is used, does not, as we have seen, belong to the figure as constructed; it corresponds in fact to the case in which the square E B is exterior to the triangle A B C, that is, when the point C is between D and B, and A in A B produced, the angular point E being in the hypothenuse A C produced. This case is evidently comprehended in the general investi

gation that has been given. The student can verify this result for himself. He should also be able to point out the limits of the problem when the negative sign is used, for the solution in some cases is impossible.

EXERCISES FOR PRACTICE.

1. Given one side of a right-angled triangle hypothenuse and the other side remaining side.

=

=

5, and the sum of the 25, to find the hypothenuse and the Ans. 13 and 12.

2. The hypothenuse of a right-angled triangle is 13, and the excess of the perpendicular above the base is 7; find the sides of the triangle. Ans. Base BC= 5, A B = 12.

3. The hypothenuse of a right-angled triangle is 13, and the sum of the base and perpendicular is 17; find the sides. Ans. 5, 12.

4. The product of the base and perpendicular of a triangle is 60 and the hypothenuse 13; what are the base and perpendicular? Ans. 12 and 5. 5. Find the side of a square inscribed in a given semicircle whose diameter is d. Ans. d√5.

6. Find a point in the diagonal of a given square (side a) from which if a perpendicular to the diagonal be drawn, the part of it intercepted between the diagonal and the side of the square may be equal to the difference of the diagonal and side.

Ans. Distance from one extremity of diagonala. 7. Having given two contiguous sides (m, n) of a parallelogram, and one of its diagonals (d), to find the other diagonal.

Ans. √(2 m2 + 2 n2 d2).

8. Given the two sides (a, b) and the line (1) which bisects the vertical

9. In a triangle, having given the segments (p, q) of the base made by a perpendicular from the vertical angle, and the ratio (m: 1) of the two sides, to find the sides.

10. In a right-angled triangle the lengths of the lines drawn from the acute angles to the points of bisection of the opposite sides are p and I, to find the sides of the triangle.

11. The radii of two circles which intersect one another are r and r'; and the distance of their centres is c; find the length of their com1

mon chord. Ans. √(r' + c+ r) (r' + c− r) (r + r'−c) (r+c−r').

с

12. Given the lengths (a, b) of two chords which cut each other at right angles in a circle, and the distance (d) of their point of intersection from the centre, to find the diameter of the circle.

Ans. (a + b2) + 2 d2 }·

13. Find the side of an equilateral triangle inscribed in a circle whose diameter is d, and that of another circumscribed about the same circle. Ans. d3 and d3.

14. Given one side AB

a of a triangle ABC, to find BC and CA, so that A C, CB, BA and a perpendicular BD on AC may be in continued geometrical progression.

15. A gentleman has a garden 60 feet long and 40 feet broad, and a walk is to be made of an equal width half round it so as to occupy half the garden; find the breadth of the walk. Ans 13 944487 feet.

16. Draw the line D PE (solution of Problem I.) so that the circles described about the triangles A EP and DCP may be equal.

Ans. M E is a fourth proportional to A M, B M, and P M. 17. The lengths of two lines that bisect the acute angles of a rightangled triangle are 40 and 50, it is required to find the sides.

Ans. 35 80737, 47·40728, and 59-41143.

18. Find the side of a regular pentagon inscribed in a circle whose diameter is d, and that of another circumscribed about the same circle. Ans. d (10- 2√√5) and d√ (5 − 2 No 5).

19. A statue eighty feet high stands on a pedestal fifty feet high, and to a spectator on the horizontal plane they subtend equal angles; find the distance of the observer from the base, the height of the eye being five feet. Ans. 99 874922 feet.

20. If from one of the angles of a rectangle a perpendicular be drawn to its diagonal, and from the point of intersection with the diagonal lines be drawn perpendicular to the sides which contain the opposite angle; then if p, p' be the lengths of the perpendiculars last drawn, and d the diagonal of the rectangle,

p3 +p'}

PLANE TRIGONOMETRY.

DEFINITIONS AND FIRST PRINCIPLES.

ART. 1. PLANE Trigonometry was restricted originally to the science by which the relations between the parts of plane triangles are established. In its extended signification, however, it is understood to comprehend the general relations of arcs and angles, and the properties which connect the sides and angles of any plane rectilineal figures.

In the varied applications of mathematics to physics, trigonometry enables us to combine the practical exactness of numerical calculations with the graphic constructions of geometry. Hence the great importance of this branch of mathematics.

The mode of measuring an angle by means of an arc of the circle whose centre is the angular point will first be pointed out, and then it will be shown how an angle may also be measured by straight lines drawn in and about the circle.

2. Let A OB be any angle; with the linear unit of length as radius, and with O as centre, describe a circle, meeting O A, O B in A and B. Produce AO to meet the circle again in E, and draw the diameter CD perpendicular to AE. Then denoting the arc A B by a, BC is called

the complement, and B E the supplement, of a; AC and ACE being, respectively, a quadrant and a semicircle. The point A may be termed the origin, and B the extremity, of the arc A B.*

In the first quadrant, it will be noticed, the complement of a is its defect from a quadrant; but in the other quadrants, it is its excess above a quadrant. And similarly for supplement of a when a is greater or less than a semicircle.

3. The right angle AOC is supposed to be divided † into 90 equal

*An augle, as defined by geometry, is less than two right angles, and hence the subtending arc of a circle whose centre is the angular point is less than a semicircumference. In trigonometry, however, the terms angle and arc have a more extended meaning. Thus the whole angular space described by the radius O B in revolving about the point O is called an angle, which may therefore be of any magnitude whatever; and, consequently, the subtending arc may consist of any number of circumferences, or parts of these.

+ This division of the right angle into 90 equal agesimal division, was that adopted by the ancients.

parts, which is called the sexAttempts have been made by