But if the length of the arc, and radius, be given, we use the formula an angle of 60°, and the radius is 10 feet; required the Ex. 2. The length of the arc of a sector is 16 yards, and the radius 124 yards; required the area of the sector. Area of sector = 16×12.5 = - 8×12·5=100 sq. yds. 242. To measure a given segment of a circle. Let ABC be the given segment; and if the centre O of the circle be not given, let it be found by (147, Part 11.). Join AO, BO; then it is clear, that the area of the segment ABC is equal to the sector AOBC diminished by the triangle AOB; that is, if the given segment be less than a semicircle. And if the segment be greater than a semicircle, the only difference will be that the triangle must E B A D be added to the sector instead of being subtracted. Hence the sector, AOBC, which is bounded by the same arc as the given segment, is to be measured by (241); and the triangle ABO by one of the usual methods; from which two results the area of the segment is readily determined. Thus, segment ABC= sector AOBC-triangle AOB, (BD being a perpendicular from B on AO, or AO produced) Or, if OE be drawn perpendicular to AB, (1) What is the unit generally adopted in the measurement of angles, 1st, theoretically, and 2ndly, practically? (2) Give the subdivisions of the smaller of these two units. (3) Describe the instrument used for measuring proposed angles, and laying down on paper angles whose numerical values are known. (4) Mention the angles which can be most readily laid down without the use of any instrument specially constructed for that purpose. (5) State the relation which exists between the circumference, and the radius, of every circle. (6) What is the ratio between the circumference of a circle, and its diameter? Is the quantity expressing that ratio a line, or an abstract number? (7) Describe the mode whereby we approximate to the area of a circle. (8) What is the relation which subsists betwixt the area of a circle and its radius? Is that relation such, that if the radius be doubled, the area will be doubled? If not, how much is the area increased? (9) Is the fraction which expresses the quotient of the area of a circle by its radius, a line, or an abstract number? (10) Write down the fraction which expresses the quotient of the area of a circle by its circumference. (11) State the difficulty we experience in finding the exact value of any circular area or circumference. Do we meet with any similar difficulty in the measurement of straight lines? (12) When using the instrument called a Protractor, shew how any error of observation, made in determining an angle in the usual way, may be diminished by one half. (13) Compare the effects of using the two numbers and 3.14159, in commonly employed for π, viz. 22 7 finding the area of a circle of 74 inches radius. Ans. The difference is 07127 sq. in. 22 [In the following examples the value of π has been taken as .] 7 (14) If the circumference of a circle be 4 poles, what is the radius? Ans. 3 yds. (15) What must be the radius of a circle, so that the circumference shall be 1 poles? Ans. 1yd. 11 in. (16) When the area of a circle is a square perch, what is its radius? Ans. 3'1024... yds. (17) The diameter of a circle is 3 yds.; what is the circumference of the circle, and what its area? (1) Ans. 11 yds. (2) Ans. 95 sq. yds. radius of the circle, sq. yds. (18) Find the measure of the of which the quadrant contains 28 Ans. 6 yds. (19) Find the circumference, approximately to 3 places of decimals, of the circle, of which the area is 16 sq. yds. Ans. 14-182 yds. (20) Given that the side of an equilateral hexagon inscribed in a circle is equal to the radius, find by what portion of the radius the semi-circumference of the circle exceeds the sum of three sides of the hexagon. Ans. (21) Find the length of the minute-hand of a clock, the extremity of which moves over an arc of 10 inches in 33 minutes. Ans. 25 in. (22) The sum of the interior angles of a regular polygon is 1800°; find how many sides it has, (See 86, Part. I.). Ans. 12. (23) The area of a circle is 29 sq. ft.; what is the area of another circle whose diameter is three-sevenths of the diameter of the former ? Ans. 5.3265 sq. ft. (24) Find the area of a segment of a circle, whose arc subtends an angle of 60° at the centre, the diameter of the circle being 10 feet. Ans. 2.27 sq.ft. (25) Prove that a cord, with its ends joined, will inclose a greater area when in the form of a circle than in the form of a square. PROBLEMS. PROB. 1. To find the numerical value of each of the angles of a right-angled isosceles triangle. By the supposition, one of the three angles is 90°, and the other two are equal to one another; but all the angles of any triangle are together equal to two right angles, therefore the two equal angles, in this case, are together equal to one right angle, that is, each of them is half a right angle, or 45°. Hence, conversely, if we have a right-angled triangle, in which one of the acute angles is known to be 45°, we may conclude, that the other angle is also 45°; and further that the sides containing the right angle are equal. This property is very useful in enabling us to find the heights of lofty buildings, or the distance between two inaccessible points, as will appear hereafter. PROB. 2. To construct angles of 60°, 30°, 15°, and 75°, without the aid of the Protractor. A Describe an equilateral triangle ABC, by (23, Part 1.); then we know that this triangle is also equiangular; and since the three angles are together equal to two right angles, or 180°, (37, Part 1.) therefore each of them is 60o. Bisect BC in D; and join AD; then BAC will be bisected by AD; and therefore BAD=30°. L Again, from DA cut off DE equal to DB, and join BE; then E C B D F ¿DEB=¿DBE; and, by Prob. 1, each =45o; therefore LABE LABD-LDBE=60°-45°=15°. Again, draw BF at right angles to AB; then Thus LEBF LABF-LABE=90°-15°-75°. <ABC=60o, <BAD=80o, ▲ABE=15°, and EBF=75o. PROB. 3. To find the area of a triangle, two sides of which are given, and contain an angle of 30°. Let ABC be the triangle, in which AB and AC are known, and <BAC=30o. B Produce BD to E, making DE equal to DB, and join AE; then LDAE=LDAB=30°; and EAB=60°. Also, since LADB-90°, and BAD=30°; .. ABD=60°. L |