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In the figure below, $\angle E = 40^\circ$ and arc $AB$, arc $BC$, and arc $CD$ all have equal length. Find the measure of $\angle ACD$, in degrees.

AdminMod2  Sep 1, 2017
 #1
avatar+88871 
+1

 

Note that  minor arcs

 

AD + AB + BC + CD  = 360°    or....put another way.....

 

AD + 3BC  = 360  →   AD  = 360 - 3BC   (1)

 

Also

 

Angle BEC  = (1/2) ( BC -  AD)      sub (1) into this

 

40  =  (1/2) [  BC - (360 - 3BC) ]       multiply through by 2  and simplify

 

80 =  4  BC - 360          add 360 to both sides

 

440  = 4  BC                divide both sides by 4

 

110°  =  BC

 

So, since AB = BC = CD.....then major arc  ACD  = 330°

And arc AD  must = 30°

And since angle ACD  is an inscribed angle intercepting this arc...

Its measure is (1/2) arc AD  = 15°

 

 

 

cool cool cool

CPhill  Sep 1, 2017
edited by CPhill  Sep 1, 2017
 #2
avatar+20009 
+1

In the figure below, \( \angle E = 40^{\circ}\) and arc AB, arc BC, and arc CD all have equal length.
Find the measure of \(\angle ACD\), in degrees.

 

 

 

1. \(\mathbf{ \angle BCE =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle BCE &=& 90^{\circ} - \frac{40^{\circ}}{2} \\ \mathbf{ \angle BCE } &\mathbf{ =}& \mathbf{ 70^{\circ}} \\ \hline \end{array} \)

 

2. \(\mathbf{ \angle BCA =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle BCE + 2\cdot \angle BCA &=& 180^{\circ} \\ \angle BCA &=& \frac{180^{\circ}- \angle BCE} {2} \\ \angle BCA &=& 90^{\circ}- \frac{\angle BCE} {2} \quad & | \quad \angle BCE = 70^{\circ} \\ \angle BCA &=& 90^{\circ}- \frac{70^{\circ}} {2} \\ \angle BCA &=& 90^{\circ}- 35^{\circ} \\ \mathbf{ \angle BCA} &\mathbf{ =}& \mathbf{ 55^{\circ}} \\ \hline \end{array} \)

 

3. \(\mathbf{ \angle ACD =\ ? }\)

\(\begin{array}{|rcll|} \hline \angle ACD &=& \angle BCE - \angle BCA \\ \angle ACD &=& 70^{\circ} - 55^{\circ} \\ \mathbf{ \angle ACD} &\mathbf{ =}& \mathbf{ 15^{\circ} } \\ \hline \end{array}\)

 

 

laugh

heureka  Sep 1, 2017

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