GMAT CLUB OLYMPICS: In the square ABCD shown above, what is the measur

Question

https://gmatclub.com/forum/download/file.php?id=62922 In the square ABCD shown above, what is the measure of an yellow angle ? A. 15° B. 20° C. 25° D. 30° E. 45° M37-25 M37-25 Untitled12.png

Bunuel · Accepted Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-25.png 
 
In the square ABCD shown above, what is the measure of an yellow angle ?

A.  15° 
B.  20° 
C.  25° 
D.  30° 
E.  45°

https://gmatclub.com/tests-beta/resources/import/img/M37-25-2.png 
 
Construct triangle BFC congruent to triangle DEC on side BC (simply put draw triangle BFC on BC the same way DEC is drawn on DC). Notice that  \angle BFC =180°-(15°+15°)=150° .
 
Since BFC and DEC are congruent, then  CE = CF . 
 
Since  \angle DCE = \angle BCF = 15° , then  \angle ECF = 90°-(\angle DCE + \angle BCF) = 60° 
 
Thus, triangle EFC is isosceles ( CE = CF ) and  \angle ECF =60° , which means that triangle EFC is equilateral and all its angles are  60° .
 
Next, since the sum of the angles around point F must be  360° , then  \angle BFE = 360° - (\angle BFC + \angle CFE)=360° - (150° + 60°)=150° 
 
Now, consider triangles EFB and CFB. In these triangles, two sides and angle between them are equal (BF is shared side, FC = FE because EFC is equilateral, and  \angle BFE = \angle BFC=150° , thus EFB and CFB are congruent. 
 
Finally, since EFB and CFB are congruent, then  \angle FBC=\angle FBE=15°  and therefore,  \angle CBE = \angle FBC+\angle FBE=30°

Answer: D

HarshaBujji · Answer

Wow, It's a famous problem;

There are 2 ways to solve it 1) Via Construction and pure geometry or 2) Via Trignometry;

The trigonometry method is a little complex; Although when I did it for the first time I used trigonometry.

Let's construct a triangle similar to DEC with BC as its base.

So <OCB =<OBC =15';
That makes <ECO = 60';

As we know OC and EC are the same; So it's isosceles triangle with an angle 60'; So It is an equilateral triangle.

Now <EOC=60';

So <EOB=150' (Sum of all angles is 360' around a point)

As OE=OB <EBO=15';

Hence <EBC=30';

IMO D

avinashiitp · Answer

From the below figure tan(a)=1/(2-tan(15) tan(15)=tan(45-30)=(1-1/3^0.5)/(1+1/3^0.5)=2-3^0.5 tan(a)=1/(2-2+3^0.5) hence a =30 degress ans D img.JPG

hnguye10 · Answer

We have ∠ECF = 90-15 = 75° = ∠EFC => EF = EC = DE = DF/2
We have ∠HEC = 30° => HC = EC/2

We have DF*HC = DC*FC (both equal two times the area of the triangle DFC)
=> 2*EC*EC/2 = BC*FC
=> EC^2 = BC*FC
=> FC/EC = EC/BC
=> triangle ECF and triangle BCE are similar
=> ∠EBC = ∠FEC = 30°

The answer is D

sivatx2 · Answer

Refer to the image attached:
 https://gmatclub.com/forum/download/file.php?mode=view&id=63288 
I have copied the triangle on the side CD to side BC.
Now the triangle CDE and BCP are congruent.

Angle EDC = Angle PCB
Angle ECD = Angle PBC
Angle CED = Angle BPC
Since CD = BC ( from square property),
We know now,

DE = CP
EC = PB

ECP = BCD – ECD – BCP
ECP = 90° – 15° – 15°
ECP = 60°

Because, we have an isosceles triangle with CP = CE, and DCE = BCF = 15° already and now with ECF = 60°, the other two sides of triangle ECP will also be 60° making it a equilateral triangle.

Since sum of all angles of the vertices of a triangle is 180°,

we can now say BPC = 150°.

Also, since sum of all the 3 angles around point P, will be 360°,
we can also say BPE = 150°.

Therefore the triangles EPB and BPC are congruent with 2 sides are same and one angle same ( with EP = CP, EPB = BPC, and PB is one of the side for each of the triangles).

Thus EBP = CBP = 15°.

Therefore, EBC = EBP + CBP = 15° + 15° = 30°.

So the correct answer choice is D

bumpbot · Answer

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GMAT CLUB OLYMPICS: In the square ABCD shown above, what is the measur

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