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Please explain how you got your answer. Thank you so much!

 

1. Triangle ABC is inscribed in circle O. We know ∠AOB = 110 degrees, ∠AOC = 120 degrees. Find the measure of ∠BAC.

2. As shown in the diagram, NM is the median of trapezoid ABCD and PQ is the median of trapezoid ABMN. We know AB = 5 and DC = 17. Find PQ. 

3. The figure shows a square in the interior of a regular hexagon. The square and regular hexagon share a common side. What is the degree measure of ∠ABC?

 Mar 3, 2021
 #1
avatar+201 
+1

2. We know that the median is the top-bottom divided by 2, so 17-5/2=6, and 6/2 is 3, and 3+5=8

 Mar 3, 2021
 #2
avatar+485 
+2

1. angle BAC=1/2 angle BOC, and BOC= 360-(110+120)=130, BAC=130/2=$\boxed{65^\circ}$

 

2.

If NM is the median of trapezoid ABCD, then that means NM is the average of the two heights, which in this case are 5 and 17. so that means NM= $\frac{5+17}2=11$

 

now to find the median of trapezoid ABMN, we have to do the same thing, take the average of the two heights, which in this case are 5 and 11. This means that PQ=$\frac{5+11}2=\boxed{8}$

 

3.

we can first start off by figuring out how many degrees are in each interior angle of a hexagon: (6-2)*180$^\circ$=720$^\circ$           $\frac{720^\circ}{6}=120^\circ$  and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have

 

$\frac{180-(120-90)}2$= angle C = $75^\circ$

and because we want to find angle ABC, we have to subtract 75$^\circ$ from 120 to get angle ABC = 120-75 = $\boxed{45^\circ}$

 Mar 3, 2021
edited by SparklingWater2  Mar 3, 2021
edited by SparklingWater2  Mar 3, 2021
 #5
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thank you sooooo much :)

Guest Mar 3, 2021
 #3
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1)   the two angles labelled encompass  110 + 120 = 230 degrees of the circle

         the other angle covers the rest      360 - 230 = 130 o

 Mar 3, 2021
 #4
avatar+485 
+1

Small error:He/she is asking for angle BAC not BOC

SparklingWater2  Mar 3, 2021

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