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).

#Geometry

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Zi Song Yeoh - 7 years, 8 months ago

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

Vikram Waradpande - 7 years, 8 months ago

Oh, I see. Indian Mathematical Olympiad?

Zi Song Yeoh - 7 years, 8 months ago

@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 - 7 years, 8 months ago

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

Jun Das - 7 years, 7 months ago

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

Vikram Waradpande - 7 years, 8 months ago

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

Kishan k - 7 years, 8 months ago

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

neha adepu - 7 years, 7 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)

Maharnab Mitra - 7 years, 8 months ago

This que is not from IMO !

Rashmi B K - 7 years, 8 months ago

I think is 180 degrees

Yen Loong - 7 years, 8 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, $A = 60°$ $B = 120°$

Rashmi B K - 7 years, 8 months ago

THANK YOU!!!

neha adepu - 7 years, 7 months ago

ANGLE A=60 ANGLE B=120

Avishkar Rajeshirke - 7 years, 8 months ago

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

Anton Than Trong - 7 years, 8 months ago

Thats Indian Maths Olympiad

Avishkar Rajeshirke - 7 years, 8 months ago

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

neha adepu - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$