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# Arc lengths

0
710
2
+166

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.

Sep 1, 2017

#1
+96032
+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°

Sep 1, 2017
edited by CPhill  Sep 1, 2017
#2
+21244
+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. $$\mathbf{ \angle ACD =\ ? }$$

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

Sep 1, 2017