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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).

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

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

Oh, I see. Indian Mathematical Olympiad?

@Zi Song Yeoh – 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.

@Pranav Arora – its called international mathematics olympiad conducted by SOF

@Zi Song Yeoh – Nope. Its a basic level contest conducted by a company.

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

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

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)

ANGLE A=60 ANGLE B=120

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, \(A = 60°\) \(B = 120°\)

THANK YOU!!!

I think is 180 degrees

This que is not from IMO !

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

Thats Indian Maths Olympiad

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

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## Comments

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

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

Log in to reply

Oh, I see. Indian Mathematical Olympiad?

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

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

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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)

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ANGLE A=60 ANGLE B=120

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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, \(A = 60°\) \(B = 120°\)

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THANK YOU!!!

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I think is 180 degrees

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This que is not from IMO !

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

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Thats Indian Maths Olympiad

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

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