Question

8) AB is parallel to FG. Angles are as shown in figure. Find and report answer as x/2.
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Answer 1

Answer:

x=130°

Step-by-step explanation:

Let transversal DE ( E is the point where angle x+10 forms) intersects AB & FG at M & N resp.

In ∆EFN, angle ENF = (x+10) - (40) = x - 30

Now, angle ENF = angle DMB = x - 30, as AB || FG.

Now, angle COD ( where O is the point of intersection of DM & CB ) = 180 - [20+(x-30)]=170-x

Now, angle CDE =x = 90+(170-x)=260-x

x = 260-x

x = 130°

hope you get it!

Answer 2

Step-by-step explanation:

Rectangle  has  and . Point  lies on  so that , point  lies on  so that , and point  lies on  so that . Segments  and  intersect  at  and , respectively. What is the value of ?





Solution 1 (Coordinate Geometry)

First, we will define point  as the origin. Then, we will find the equations of the following three lines: , , and . The slopes of these lines are , , and , respectively. Next, we will find the equations of , , and . They are as follows:After drawing in altitudes to  from , , and , we see that  because of similar triangles, and so we only need to find the x-coordinates of  and .Finding the intersections of  and , and  and  gives the x-coordinates of  and  to be  and . This means that . Now we can find 

Solution 2 (Similar Triangles)



Extend  to intersect  at . Letting , we have that

Then, notice that  and . Thus, we see thatandThus, we see that

Solution 3 (Answer Choices)

Since the opposite sides of a rectangle are parallel and    due to vertical angles,   . Furthermore, the ratio between the side lengths of the two triangles is   . Labeling    and   , we see that  turns out to be equal to . Since the denominator of  must now be a multiple of 7, the only possible solution in the answer choices is .

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