Suppose a spider moves along the edge of a circular web at a distance of 3 cm from the center.?(a) If the spider begins on the far right side of the web and creeps counterclockwise until it reaches the top of the web, approximately how far does it travel? Distance: ______units(b) If the spider crawls along the edge of the web a distance of 1 cm, approximately what is the angle formed by the line segment from the center of the web to the spider's starting point and the line segment from the center of the web to the spider's finishing point? Angle: _______degrees
Accepted Solution
A:
Answer:a) d = 4,712 cmb) ∠ 108,52 ⁰ Step-by-step explanation:The crcular path of the spider has radius = 3 cmIf the spider moves from the fa right side of the circle its start poin of the movement is P ( 3 , 0 ) (assuming the circle is at the center of the coordinate system. And the top of the web (going counterclockwise ) is the poin ( 0 , 3 )So far the spider has traveled 1/4 of the circleThe lenght of the circle isL = 2*π*rThe traveled distance for the spider (d) is d = (1/4 )* 2*π*r ⇒ d = (1/2)*3,1416*3 ⇒ d = 4,712 cmAnd now spider go ahead a new distance of approximately 1 more cmTherefore the spider went a total of 5.712 cmNow we know that πrad = 180⁰ then 1 rad = 180/π ⇒ 1 rad = 57⁰rad = lenght of arc/radius so 5,712/3 = 1.904 radBy rule of three 1 rad ⇒ 57⁰ 1.904 rad ⇒ ? xx = 57 * 1.904 ⇒ 108.52 ⁰So the ∠ 108,52 ⁰