money in (A) which is put out to interest. Therefore the above result must hold for the varying quantity in (B).

Here P = a; and M = b, when n = t,, the unit of time being

arbitrary; and the constant ratio

1

Therefore

m + 1°

and if x is the required value at the time t,

(1),

(2).

1850. (A). Express the number of numbers less than a given number which are prime to it, in terms of the given number and its prime factors.

(B). Shew that the sum of these numbers is equal to half the product of the number of them into the given number.

From (4) we find that the number of numbers prime to and

(1). Let N = a". Then the numbers < N and not prime to

therefore sum of numbers < N and prime to N

then, the numbers divisible by a and < N are

Hence, observing that each of the two former sets include the latter, the sum of all numbers less than N and not prime

therefore sum of the numbers <N and prime to it

1851.

TRIGONOMETRY.

(A). Compare the magnitudes of two angles which contain the same number of French and English degrees respectively.

(B). Divide an angle which contains n degrees into two parts, one of which contains as many English minutes as the other does French.

From (4) we get, that if A and B be two angles of which the former contains as many English degrees as the latter does French grades,

A

10

If A contained as many English minutes as B contains French, the above formula would have to be modified into

Let now A, B be the two required parts of the given angle, expressed in English degrees. Then we have

(B) Construct the angle whose tangent is 3−√2.

In the investigation of (4), we prove that

1

cos 45°

√/2

Take any finite straight line

right angles to AB and 3AB.

AB (fig. 28). Draw BC at

Make BD = AB and join AD. Cut off from CB, CE-AD, and join AE. BAE shall be the angle required.

For, since BD = AB, the 4 BAD = 45°;

1851.

(4), Prove that sin (A+B) = sin A cos B + cos A sin B, and deduce a similar expression for cos (A+B).

1848.

(A). Express sin2A in terms of tan A.

(B). Given tan 4 =

From (4), we have

2 √√3, find sin A, and thence A.

2 tan A

1 + tan2 A '

therefore

1850.

(A). If A + B + C = 180°, prove that

tan A+ tan B + tan C =tan A tan В tan C.

(B). If a, ß, y, denote the distances from the angular points of a triangle, to the points of contact of the inscribed circle, shew that the radius of the inscribed circle

аву
a+B+Y

Here (B) is a direct application of the formula proved in (4).

If O (fig. 29) is the centre of an inscribed circle, the lines OA, OB, OC, bisect angles A, B, C.