Question 1096676
By how much does the arc intercepted by a central angle of 38 degrees exceed the chord intercepted by the same angle on a circle of radius 43 ft.?


Please help me :(
<pre>Let central angle be O, radii AO and BO, and intercepted chord & intercepted arc, AB
&#8737A = 38<sup>o</sup>
Length of arc AB: {{{matrix(1,16, 38/360, "*", 2pi, "*", r, "=", 19/180, "*", 2(43), "*", pi, "=", 19pi(43)/90, "=", 817pi/90, feet)}}}

&#8737;A and &#8737;B = {{{matrix(1,5, (180 - 38)/2, "=", 142/2, "=", 71^o)}}} 
Draw an altitude from vertex O to AB, and name it C
cos &#8737;A = {{{matrix(1,3, A/H, "=", AC/AO)}}}
{{{matrix(1,3, cos (71^o), "=", AC/43)}}}
{{{matrix(1,5, AC, "=", 43, "*", cos (71^o))}}} --------- Cross-multiplying
AC = 14
Since the altitude from vertex O, or OC BISECTS chord AB, AC = {{{matrix(1,3, (1/2), of, AB)}}}. Therefore, the INTERCEPTED CHORD or {{{matrix(1,6, AB, "=", 2(14), or, 28, feet)}}}
Arc AB exceeds chord AB by {{{highlight_green(matrix(1,9, 817pi/90 - 28, or, 28.52 - 28, "=", .52, "feet,", or, 6.24, inches))}}}.