5. In any triangle the square on the side subtending an acute angle is less than the squares on the sides containing that angle by twice the rectangle contained by either of those sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle and the acute angle. A, B are two given points, and CD a given straight line; in CD find a point P such that the difference between the squares on AP, BP shall be equal to twice the square on AB. 6. Make a square equal in area to a given equilateral triangle. 7. Draw a tangent to a given circle from a given point outside it. 8. The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles. If the opposite sides of such a quadrilateral be produced to meet in P and Q, and about the triangles so formed outside the quadrilateral circles be described intersecting again in R, show that P, R, Q will be in one straight line. 9. Describe a circle which shall touch one side of a given triangle and the other two sides produced. IO. Describe an isosceles triangle having each angle at the base double of the vertical angle. Construct a quadrilateral whose angles, taken in order, are in the proportion 1:2:3:4, and whose longest side is double of that opposite to it. II. If the vertical angle of a triangle be bisected by a straight line which cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another. The angle A of a triangle ABC is bisected by AD, which cuts the base at D, and O is the middle point of BC; show that OD bears the same ratio to OB that the difference of the sides bears to their sum. 12. In any right-angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarlydescribed figures on the sides containing the right angle. I. (Up to and including the Binomial Theorem, the theory and use of Logarithms.) and divide 2. Simplify 4. [N.B.-Great importance will be attached to accuracy.] Prove that (4x − 1 )3 + (2x − 3)3 +6(4x − 1)(2x − 3)(3x − 2) = 8(3x -- 2)3 ; (x − 1)8 + (x − 1 )4+1 by (x-1)* + (x − 1 )2 + 1. 2(x2 - 1)2 (x2+2)(x2-16) and resolve into factors each of the expressions 5. II. ALGEBRA. x2-26x+168 and x+x22+14. 3. Show that the sum of the squares of 5 consecutive numbers is divisible by 5. (x-4) (x − 1)2+(x+4)(x+1)2 ̧ Find the highest common factor of Solve the equations 4x+7x2+16 and 4x4+12x2+9x2 – 16. (ii.) (i.) 3+3 = x2-1 (x-1) 3 2 35 + 3 x X-I (ii.) 16x2 3612 4x 6y 3 5 ax b b = 1. 6. Transform 5363433 from the septenary scale to the denary. Multiply together the two numbers 4535 and 246 in the septenary scale. 7 Solve the equations (i.) = - ax 33 = 4, Sx3 − xy+y3 = 4, 8. A courtyard, containing 1300 square feet, is paved with 50 stones of one size and 195 of another. If 10 stones of the first size and 7 of the second make together 100 square feet, find the size of each stone. 9. By Mathematical Induction, (i.) prove the formula for the summation of a Geometrical Progression ; (ii.) show that the series (15+17) + (25 +27)+(35+37) + etc. (to » terms) n* (n+1)4 8 = 10. Find an infinite geometrical progression whose first term is I, and in which each term is twice the sum of all the terms that follow it. Define Harmonical Progression; and place two terms at each end of the H.P. 15, 20, 30. II. Find for what value of the number of combinations of 2m things taken at a time is greatest. Show that there are two terms in the expansion of (1 - x) all the others, when x = 19. greater than 12. When the logarithms of two adjacent numbers are known, explain how to determine the logarithm of a number lying between them. Find the tenth root of 4325626. If £35627. 19s. 4d. be placed at compound interest at the rate of 34 per cent. per annum, what will be the amount at the end of 30 years? III. PLANE TRIGONOMETRY AND MENSURATION. I. The tangent of an angle is 2'4. Find the values of the cosecant of the angle, the cosecant of the complement of half the angle, and the cosecant of the supplement of double the angle. 2. (Including the Solution of Triangles.) [N.B.-Great importance will be attached to accuracy.] Establish the identities (i.) (cosec A+ sec A)2- (cos A+ sin A)2 (ii.) 2 = (cot A cosec A+tan 4 sec A)(sin A + cos A)(1 + sin A cos A). 2 cos2A I+tan A sin2A 3. Simplify and prove that + sin 24 + 4. Prove that = 2. I+tan 0+tan20√1+cot + cot20 - sec @ cosec = I. 5. In a plane triangle show that the sines of the angles are to one another as the opposite sides. Show that a cos A cos 24 sin 4(B - C)+b cos B cos 2B sin 4(C – A) 6. In the ambiguous case in the solution of plane triangles, if a and b are the two given sides a<b, A the given angle, and C1, C2 the two values of the angles opposite the third side, prove that cos A cos B G2+ Ca = sin . sin G-C2, 2 2 taking the smaller of the two values of B. 7. In any triangle ABC, if r, ra, ro, re, are the radii of the inscribed and of the three escribed circles respectively, and pa, Po, pe, the perpendiculars from the angles on the sides a, b, c, prove that I I I I = + + ; 1 Pa Po Pc I I I and if the points where the perpendiculars meet the sides be joined, the perimeter of the triangle so formed is 2a sin B sin C. 8. Find the values of the angle C in the triangle ABC from the following data: b = 785, c = 1083, B = 37° 11′ 45′′. 9. Adapt to logarithmic computation the expression (x+√x2-1)+(x − √x2 − 1)3. IO. A railway tunnel consists of a hollow semi-cylindrical top, terminated below in a trough with slanting sides and flat base. The radius of the former being 12 feet, the base and height of the latter being 20 and 18 feet respectively, and the length of the tunnel 1200 yards, find the cost of facing the sides and roof with brick at Is. 6d. per square foot (π = 22). II. Within a hollow sphere of 1 foot radius is placed a right prism, the ends of which are equilateral triangles. The side of one of these being I foot in length, and the surface of the sphere being in contact with all the six angular points of the prism, find in cubic inches the volume of the latter. |