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# some math

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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
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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
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Excellent, tertre!!!

CPhill  Feb 20, 2019
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Thank you, CPhill!

tertre  Feb 20, 2019
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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 $$\mathbf{23^\circ}$$

Feb 20, 2019
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Very nice, Heureka!!!!

CPhill  Feb 20, 2019
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Thank you, CPhill !

heureka  Feb 21, 2019
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Heureka, where was it stipulated that ABC is a right angle?  I'm going crazy trying to find that.

Feb 21, 2019
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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°

CPhill  Feb 21, 2019
edited by CPhill  Feb 21, 2019
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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
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Yep...C can be located  anywhere on the circle [ except at A or B ]  and angle ACB will be a right angle !!!

CPhill  Feb 21, 2019