论文部分内容阅读
(1)联直线OB(图1),O为圆心,A为切点,∴△OAB为直角三角形,OA为半径,即OA==10/2=5。∵OB平分∠ABC(角的两边与圆相切),∴∠ OBA=55°/2=27°30′。由直角三角形OAB得:AB=OA·ctg 27°30′=5×1.921=9.605。由图可知:x=40+2AB+2×5=40+2×9.605+10=69.21。答:x为69.21毫米(2)联O_1O_2、OO_1、OO_2三线,作OA⊥O_1O_2(图2),∠O_1OO_2=360°/5=72°,∵OO_1=OO_2 ∴△OO_1O_2为等腰三角形,则O_1A=O_2A, ∠O_1OA=36°,(等腰△的高平分底边与顶角)。
(1) joint line OB (Figure 1), O is the center of the circle, A is the tangent point, △ △ OAB is a right triangle, OA radius, that is, OA == 10/2 = ∵OB bisect ∠ABC (both sides of the circle tangent to the circle), ∴∠ OBA = 55 ° / 2 = 27 ° 30 ’. From the right triangle OAB: AB = OA · ctg 27 ° 30 ’= 5 × 1.921 = 9.605. The figure shows: x = 40 + 2AB + 2 × 5 = 40 + 2 × 9.605 + 10 = 69.21. A: x is 69.21 mm (2) O_1O_2, OO_1 and OO_2 are the isosceles triangles for OA ⊥ O_1O_2 (Fig.2), ∠O_1OO_2 = 360 ° / 5 = 72 °, ∵OO_1 = OO_2 ∴ △ OO_1O_2 O_1A = O_2A, ∠O_1OA = 36 ° (high bisecting base and apex of isosceles triangle).