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1. In the figure shown, \(ABCD \) is a square and \(\triangle CDE\) is equilateral. What is the degree measure of \(\angle CBE\)?

 

2. A triangle whose side lengths are whole numbers has one side which measures 25 inches and a perimeter of 80 inches. What is the fewest number of inches that can be the length of one of the remaining sides?

 

3.

In the diagram below, \(AM=BM=CM\) and \(\angle BMC+\angle A = 201^\circ\) Find  \(\angle B\) in degrees.

 Feb 20, 2019
 #1
avatar+4296 
+3

1. Angle C measures 60+90=150 degrees. Since it forms a triangle, the answer is just 180-150=30/2=15 degrees.

 Feb 20, 2019
 #3
avatar+101870 
+2

Excellent, tertre!!!

 

 

cool cool cool

CPhill  Feb 20, 2019
 #4
avatar+4296 
+2

Thank you, CPhill! smiley

tertre  Feb 20, 2019
 #2
avatar+22554 
+5

3.

In the diagram below,  \(AM=BM=CM ={\color{red}\mathbf{r} } \) and  \(\angle BMC+\angle A = 201^\circ\)

Find  \(\angle B\)  in degrees.

 

\(\text{$\triangle ABC$ is a right-angled triangle, so $\angle A = 90^\circ - \angle B$ } \\ \text{$\angle CMA = 2\angle B$ }\\ \text{$\angle BMC = 180^\circ - \angle CMA$ }\)

 

\(\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 201^\circ \quad | \quad BMC = 180^\circ - \angle CMA \\ 180^\circ - \angle CMA+\angle A &=& 201^\circ \quad | \quad CMA = 2\angle B \\ 180^\circ - 2\angle B+\angle A &=& 201^\circ \quad | \quad A = 90^\circ - \angle B \\ 180^\circ - 2\angle B+90^\circ - \angle B &=& 201^\circ \\ 270^\circ - 3\angle B &=& 201^\circ \\ 3\angle B &=& 270^\circ- 201^\circ \\ 3\angle B &=& 69^\circ \\ \mathbf{\angle B} & \mathbf{=} & \mathbf{23^\circ} \\ \hline \end{array}\)

 

 

\(\angle B\)  in degrees is \(\mathbf{23^\circ}\)

 

laugh

 Feb 20, 2019
 #5
avatar+101870 
+3

Very nice, Heureka!!!!

 

 

cool cool  cool

CPhill  Feb 20, 2019
 #6
avatar+22554 
+3

Thank you, CPhill !

 

laugh

heureka  Feb 21, 2019
 #7
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0

Heureka, where was it stipulated that ABC is a right angle?  I'm going crazy trying to find that.   

 Feb 21, 2019
 #8
avatar+101870 
+1

In Heureka's diagram, angle ACB intercepts  diameter  AB

 

So....the arc that it intercepts = 180°

 

But...since angle ACB is an inscribed angle interceptng this arc, it has 1/2 of its measure = 90°

 

So...remember....any inscribed angle intercepting a diameter = 90°

 

 

cool cool cool

CPhill  Feb 21, 2019
edited by CPhill  Feb 21, 2019
 #9
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If we say that straight lines AC and BC in the diagram are elastic, and I move point C anywhere around the circle, angle ACB would always be 90o?  Have I grasped it?  I can't remember being taught that, but of course school was a long time ago for me.  Thanks.  Ron

 Feb 21, 2019
edited by Guest  Feb 21, 2019
 #10
avatar+101870 
0

Yep...C can be located  anywhere on the circle [ except at A or B ]  and angle ACB will be a right angle !!!

 

 

cool cool cool

CPhill  Feb 21, 2019

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