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What is the value of x in the figure above?

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by AAPL » Fri Dec 08, 2017 10:13 am
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In the figure above O is the center of the circle and$$\angle A$$is a right angle. If AE=AB, and$$∠EAB=90º$$, and BC=CD=DE. What is the value of x?

A. 150
B. 152
C. 160
D. 164
E. 172

The OA is A.

Experts, can you assist me with this PS question please. I have no idea about how can I solve it.

I think that I can starting assuming that EB is equal to the diameter of the circle but I don't know how can I prove it.
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by EconomistGMATTutor » Sun Dec 10, 2017 12:41 pm
Hello AAPL.

Let's take a look at your question.

For the Inscribed angle theorem, we have that m(EOB)=2*m(EAB)=2*90º=180º.

Now, the triangle EAB is equilateral and it has a right angle, so the m(AEB)=m(EBA)=45º. If we consider the line AO, then m(EOA)=m(BOA)=90º.

So, xº=90+yº, where yº=m(COB).

Using again the Inscribed angle theorem we have that 180º=m(EOB)=2*m(EDB), it implies that m(EDB)=90º.

Now, the triangles DCB and DCE are isosceles, and m(DEC)=m(DCE)=m(CDB)=m(DBC)=Rº.

If we look the triangle DCE we can get that Rº+Rº+Rº+m(BDE)=180º, it implies that Rº=30º.

So, m(CDE)=120º.

Finally, using the Inscribed angle theorem, we have that the complementary angle in COE satisfy that 90º+90º+yº=2*m(CDE)=2*120=240º. That is to say, yº=60º.

So we get xº= 90º + yº = 90º+60º=150º.

The correct answer is A.

I hope this explanation may help you.

You have to take care when you read it.

I'm available if you'd like a follow-up.

Regards.
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by GMATWisdom » Sun Dec 10, 2017 2:54 pm
AAPL wrote:Image

In the figure above O is the center of the circle and$$\angle A$$is a right angle. If AE=AB, and$$∠EAB=90º$$, and BC=CD=DE. What is the value of x?

A. 150
B. 152
C. 160
D. 164
E. 172

The OA is A.

Experts, can you assist me with this PS question please. I have no idea about how can I solve it.

I think that I can starting assuming that EB is equal to the diameter of the circle but I don't know how can I prove it.
Image

The angle formed by joining the end points of a diameter to any point on the circle is 90 degrees.

Since, Angle EAB = 90 degrees, so EB is the diameter.

Hence OE = OB = OA = radius of the circle.

EB = AB (given).

Hence triangles EOA and BOA are congruent because all 3 corresponding sides of both the triangles are equal.
hence angle EOA = angle AOB
Since angle EOA + angle AOB = Angle EOB = 180 degree so Angle 2AOB = 180 degrees.
So angle AOB = 90 degrees

Similarly, OE = OD = OC = OB = radius of the circle. ED = DC = CB (given). So triangles EOD, DOC and COB are also congruent as they have corresponding sides equal.
Therfore angle BOC = angle COD = angle DOE
angle BOC + angle COD + angle DOE = 180 degrees
Therefore 3BOC = 180
or angle BOC = 60 degrees
x = Angle AOB + AngleBOC = 90 + 60 = 150 degrees.

Hence A
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