Find the measure of the arc or central angle indicated. Assume that lines which appear to be diameters are actual diameters.

Mathematics

1 answer:

vfiekz [6]2 years ago

8 0

The measure of the Central Angle indicated is; 45°

<h3>How to find the central angle of a circle?</h3>

The central angle will be the angle subtended at the center by arc FE and arc CB.

Now, we see that angle ∠FOB = 135° and since we know that sum of angles on a straight line is 180°, then we can say that;

Central Angle = 180° - 135° = 45°

Angle subtended by Arc CD at the center is;

θ = 180 - (81 + 45)

θ = 54°

Read more about Central Angle at; brainly.com/question/17074363

#SPJ1

You might be interested in

5 Exam-style ABCD is a kite.

Mnenie [13.5K]

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

$y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}$