# Problem from IMO ABCD is a rhombus in which the altitude from d bisects AB. AE=EB. Therefore, angle A and angle B respectively are (of how many degrees). 7 years, 9 months ago

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I don't think this is from IMO (International Mathematical Olympiad).

- 7 years, 9 months ago

This is from an exam conducted here in India by a private company.

- 7 years, 9 months ago

Oh, I see. Indian Mathematical Olympiad?

- 7 years, 9 months ago

No way, I doubt they ask such easy questions. As Vikram said, it must be a contest by some private organisation.

If the private organisations use names such as "IMO" and mislead the students, then its a very bad tactic to promote themselves.

- 7 years, 9 months ago

its called international mathematics olympiad conducted by SOF

- 7 years, 9 months ago

Nope. Its a basic level contest conducted by a company.

- 7 years, 9 months ago

Hey guys,I think she is talking of SOF IMO and not the Great and the one you all are thinking IMO!!!!

- 7 years, 9 months ago

YOU,RE RIGHT!. It is a problem from the work book.

- 7 years, 9 months ago

Consider the sides of the rhombus to be of length $x$, i.e., $AD=x$. So, $AE= \frac{x}{2}$.

Let $\angle A= \theta$. So, $cos \theta = \frac{AE}{AD} =\frac{1}{2}$

$\implies \theta =60^o$ $\implies \angle A=60^o$ and $\angle B=180^o-60^o=120^o$(As they are interior opposite angles)

- 7 years, 9 months ago

This que is not from IMO !

- 7 years, 9 months ago

I think is 180 degrees

- 7 years, 9 months ago

Consider- $DA = x$

$AE = x/2$

$EB = x/2$

$DE = x^{2} - (x/2)^{2}$ $= \sqrt{3}x/2$

$DB = x$ (Pythagoras Theorem)

$DA = x , DB = x , AB = x$

$ADB$ is an Equilateral triangle

Thus, $A = 60°$ $B = 120°$

- 7 years, 9 months ago

THANK YOU!!!

- 7 years, 9 months ago

ANGLE A=60 ANGLE B=120

- 7 years, 9 months ago

If this problem is from IMO, please tell me what year and what question.

- 7 years, 9 months ago

- 7 years, 9 months ago

this problem is not from the main exam, it is from the workbook

- 7 years, 9 months ago