A CANDIDATE, PLYMOUTH.--You will see in our present issue the position of your town as a competition centre for Lower Division Clerks. T. G., GLASGOW.-Your city has not hitherto flourished in the successful Civil Service lists, we think in consequence of many local openings for junior clerks. We know no valid reason why Glasgow men should not be more fortunate. T. C., DUBLIN.-We only give your handwriting 60 per cent. Those who gave you 80 know little of either the Civil Service style or standard. H., ARMAGH.-Messrs. Longmans & Co. are the publishers of the "Civil Service Arithmetic." We believe it is in its sixteenth edition. You can get a copy at any Irish National School. JOHN S., PRINCES-STREET, EDinburgh. - You will find in our present issue valuable information as to entrance into the Excise. B. P., PROSPECT-ROW, WOOLWICH.-Thanks for your letter. We believe everyone who understands the English language, knows the mean ing of "Competitor." We are perfectly satisfied with the success of our first-born. We cannot communicate with correspondents by letter, nor can we return books sent for review. S. KING, CLONMAINE, LOUGHALL, ARMAGH.-We regret we cannot give solutions of difficult questions given years ago at Civil Service examinations. Apply to some neighbouring schoolmaster, or send us a copy of the question and we shall look at it. H. L., HIGH-STREET, STRATFORD, E.-(1) We are not aware that the attempts to shorten Euclid in France and Germany have been successful. Similar attempts have failed in this country. Dr. Morrell has written a book on the subject, we believe, in Cassell's Series; Mr. Wilson also attempted the simplification of Euclid. (2) We shall have lessons in as many subjects as we have space for, including Latin and German. In our next we hope to give Cæsar, Book I. J. A. C.-Thanks; you are right. See similar paper in this number. =sin (90°- A) cos (−B)+sin (— B) cos (90° — A). But sin (90°-4)=008 A, cos (B)=cos B, sin (B)=sin B and cos (90°-A)=sin A. .. cos (A+B)=cos A cos B-sin A sin B. Ans. π π Now substitute the given values -a and for a and ß respectively. =sin' (-a)+sin' (-3)+{cos [-(a+3)]-cos (a-3)} sin [−(a+ß)] [cos a-sin a]'+ [cos B-sin ẞ]+sin (a+B) cos (a+B)-cos (a-3) cos (a+ß) +sin (a+B) cos (a+ẞ)—cos (a−ẞ) cos (a+ß) =}{2-sin 2 a-sin 2 B}+sin (a+B) cos (a+ẞ)—cos (a-ẞ) cos (a+ẞ) 2 +sin (a+ß) cos (a+3)-cos a+sin3 B •.• sin3 a+cos3 a=1, sin 2 a+sin 2 ß=2 sin (a+ß) cos (a−ẞ) = sin' a+sin' ẞ-{cos (a-3)-cos (a+b)} sin (a+ß) = sin' a+sin' ẞ-2 sin a sin ẞ sin (a+ß) ...(cos (a-3)-cos (a+3)}=-2 sin a sin 3. Q. E. D. 4. The area of any triangle is equal to half the product of two of its sides into the natural sine of their contained angle. which is an expression for ('D in terms of the two given sides a, b, and cosine of half-given angle C. This answers the question; the student, however, who is not working against time should verify this result, which he can do by substituting the following values: a=9, b=7, c=14, then CD should be found =4. It is also a useful exercise to adapt this expression for the value of CD to logarithmic computation, as follows: 4CD2=2a2+262 — c3. |