If A+B+C= 90°, prove that cosec A cosec B cosec C-cot B tan C-cot C tan B-cot C tan A -cot A tan C-cot A tan B – cot B tan A = 2. The order of the letters denoting the direction in which a straight line is measured, AB+ BA=0 and for any three points A, B, C in a straight line AB+BC= AC. Then for any magnitude of the angle POM, and this will be true however the figure be drawn as in figs. 12 and 13. The expression cosec A cosec B cosec C- tan A (cot B+ cot C) —... =cosec A cosec B cosec C – sin A cosec B cosec C—..... =cosec A cosec B cosec C (1 - sin A - sin'B - sin3 C) = cosec A cosec B cosec C (cos3A – sin3 B — sin2 C) =cosec A cosec B cosec C { cos (A – B) cos (A + B) — sin*C} =cosec A cosec B {cos (A – B) — cos (A + B)} x. Prove that the sides of a triangle are proportional to the sines of the opposite angles. Shew that if the squares of the sides of a triangle are in arithmetical progression, the tangents of the angles are in harmonical progression. xi. Find the radius of the inscribed circle of a triangle in terms of the angles and one side. 3 red If R, r, r, r, are the radii of the circumscribed, inscribed, and escribed circles of a triangle, prove that For r1 + r2+ r2 − r = 4R. 2 3 COS COS 2 2 2 2 2 + b + A B COS 2 B C 2 COS a sin COS xii. Prove that sin lies between 0 and 0 circular measure of an angle between 0 and 90°. If a triangle be solved from the observed parts C=75°, b=2, a = √(6), shew that an error of 10" in the value of C would cause an error of about 3"66 in the calculated value of B. 1. FIND an expression for the magnitude of the resultant of any number of given forces acting at a point in given directions in a plane. Prove that the resultant of forces 7, 1, 1 and 3 acting from one angle of a regular pentagon towards the other angles taken in order is √(71). 2. State the conditions for the equilibrium of any number of forces acting upon a body in one plane and prove that they are necessary and sufficient. If six forces acting on a body be completely represented, three by the sides of a triangle taken in order, and three by the sides of the triangle formed by joining the middle points of the sides of the original triangle, prove that they will be in equilibrium if the parallel forces act in the same direction and the scale on which the first three forces are represented. be four times as large as that on which the last three are represented. Each set of three forces will be equivalent to a couple, and the two couples will balance. 3. If a system of parallel forces act at given points in a plane, find the distance of the centre of the system from a given straight line in that plane. A triangular lamina is supported at its three angular points and a weight equal to that of the triangle is placed upon it; find the position of the weight if the pressures on the points of support are proportional to 4a+b+c, a + 4h+c, a+b+4c, where a, b, c are the lengths of the sides of the triangle. The resultant of a+b+c acting at each angular point is 3 (a+b+c) acting at the centre of inertia of the triangle, and the resultant of 3a, 3b, 3c acting at the angular points is 3 (a+b+c) acting at the centre of the inscribed circle. Therefore the weight must be placed at the centre of the inscribed circle. 4. Describe the common steelyard, and shew that the distances between the graduations are proportional to the differences of the weights to which they belong. In a weighing machine constructed on the principle of the common steelyard the pounds are read off by graduations reaching from 0 to 14, and the stones by weights hung at the end of the arm; if the weight corresponding to one stone be 7 oz., the moveable weight lb. and the length of the arm one foot, prove that the distances between the graduations are in. If ACB be the beam, C the fulcrum, and O the zero of graduations, then at O and at B will have the same moment about Casat B; therefore |