PROBLEM II.

In a right-angled triangle, having given the hypothenuse, and the sum of the base and perpendicular, to find these two sides.

Put BC α = 5, BA = x, AC = y, and the sum of the base and perpendiculars = 7.

By completing the square y2 - sy + ‡s2 = }a2 −45o,

In a rectangle, having given the diagonal and perimeter, to find

the sides.

Let ABCD be the proposed rectangle.

Put AC d 10, the perimeter = 2a = 28,

= =

or AB +BC= a = 14: also put AB = x,

and BC= y.

PROBLEM IV.

Having given the base and perpendicular of a triangle, to find the side of an inscribed square.

Let ABC be the triangle, and HEFG the inscribed square.

Put

or,

ab

bx = ax,

ab

x=

a+b

= the side of the inscribed square;

which, therefore, depends only on the base and altitude of the triangle.

PROBLEM V.

In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within, on the three sides to determine the sides of the triangle.

F

E

Let ABC be an equilateral triangle: DG, DE and DF the given per pendiculars let fall from D on the sides. Draw DA, DB, DC, to the vertices of the angles, and let fall the perpendicular CH on the base. Let DG a, DE=b, and DF=e: put one of the equal sides AB = 2x; hence, AH = x, and CHVAC - AH2 = √4x2 -x2 = = √3x2 = x √3.

=

A

HG

B

Now, since the area of a triangle is equal to half its

base into the altitude, (B. IV., P. 6),

ABX CH = x x x √3 = x2 √3 = triangle ACB,

But the last three triangles make up, and are consequently equal to, the first;

hence, x2 √3 = = ax + bx + cx = x (a + b + c) ;

REMARK. Since the perpendicular CH is equal to x√3, it is consequently equal a + b + c: that is, the perpendic ular let fall from either angle of an equilateral triangle on the opposite side, is equal to the sum of the three perpendiculars let fall from any point within the triangle on the sides respectively.

PROBLEM VI.—In a right-angled triangle, having given the base and the difference between the hypothenuse and perpendicular, to find the sides.

PROBLEM VII.-In a right-angled triangle, having given the hypothenuse, and the difference between the base and perpendicular, to determine the triangle.

PROBLEM VIII.-Having given the area of a rectangle inscribed in a given triangle; to determine the sides of the rectangle.

PROBLEM IX.—In a triangle, having given the ratio of the two sides, together with both the segments of the base made by a perpendicular from the vertical angle; to determine the triangle.

PROBLEM X.-In a triangle, having given the base, the sum of the two other sides, and the length of a line drawn from the vertical angle to the middle of the base; to find the sides of the triangle.

PROBLEM XI.-In a triangle, having given the two sides about the vertical angle, together with the line bisecting that angle and terminating in the base to find the base.

PROBLEM XII. To determine a right-angled triangle, having given the lengths of two lines drawn from the acute angles to the middle of the opposite sides.

PROBLEM XIII. To determine a right-angled triangle, having given the perimeter and the radius of the inscribed circle.

PROBLEM XIV.-To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides. PROBLEM XV.-To determine a right-angled triangle, having given the hypothenuse, and the side of the inscribed

square.

PROBLEM XVI.-To determine the radii of three equal circles, described within and tangent to, a given circle, and also tangent to each other.

PROBLEM XVII.—In a right-angle triangle, having given the perimeter and the perpendicular let fall from the right angle on the hypothenuse, to determine the triangle.

PROBLEM XVIII.-To determine a right-angled triangle, having given the hypothenuse and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle.

PROBLEM XIX.-To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides.

PROBLEM XX. To determine a triangle, having given the base, the perpendicular, and the rectangle of the two sides.

PROBLEM XXI.-To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides.

PROBLEM XXII.—In a triangle, having given the three sides, to find the radius of the inscribed circle.

PROBLEM XXIII.—To determine a right-angled triangle, having given the side of the inscribed square, and the radius of the inscribed circle.

PROBLEM XXIV. To determine a right-angled triangle, having given the hypothenuse and radius of the inscribed circle.

PROBLEM XXV.-To determine a triangle, having given the base, the line bisecting the vertical angle, and the diam eter of the circumscribing circle.

PLANE TRIGONOMETRY.

1

INTRODUCTION.

OF LOGARITHMS.

1. The logarithm of a number is the exponent of the power to which it is necessary to raise a fixed number, in order to produce the first number.

This fixed number is called the base of the system, and may be any number except 1: in the common system, 10 is assumed as the base.

2. If we form those powers of 10, which are denoted by entire exponents, we shall have

10°=1

101 = 10,

102 = 100,

103=1000
101=10000, &c., &c.,

From the above table, it is plain, that 0, 1, 2, 3, 4, &c., are respectively the logarithms of 1, 10, 100, 1000, 10000, &c.; we also see, that the logarithm of any number be tween 1 and 10, is greater than 0 and less than 1: thus, log 2=0.301030.

The logarithm of any number greater than 10, and less than 100, is greater than 1 and less than 2: thus,

log 501.698970.

The logarithm of any number greater than 100, and less than 1000, is greater than 2 and less than 3: thus, log 1262.100371, &c.