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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.

sudsw12 Feb 20, 2019

#1**+3 **

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

tertre Feb 20, 2019

#2**+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 \(\mathbf{23^\circ}\)

heureka Feb 20, 2019

#7**0 **

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

Guest Feb 21, 2019

#9**0 **

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 90^{o}? Have I grasped it? I can't remember being taught that, but of course school was a long time ago for me. Thanks. Ron

Guest Feb 21, 2019

edited by
Guest
Feb 21, 2019