Triangle ABC is a isosceles right-angled triangle.A = C = 45 degrees.Therefore, AB = BC = Sq.root(2)

Area of triangle ABC = 1/2 * Sq.root(2) * Sq.root(2)
= 1/2 * 2
= 1

In Triangle DEF , angle F= 30 degrees.

Sin 30 = 1/2 =DE/2

DE = 1

Also, Cos 30 = Sq.root(3)/2 = EF/2

EF = Sq.root(3) = 1.732

This implies area of triangle DEF = 1/2 * 1 * 1.732
= .866

Clearly,Area of ABC is greater than Area of DEF.

Right triangle 1:1:root{2} and 1:root{3}:2....the one with more symmetry will have the higher area.

Do they ask questions based on sin and cos formulas in GRE ?

No, definitely.

Regards

Trigonometry is not tested on GRE or GMAT.
This means there is always another way to do the question..

Here we have two types of triangles..
45-45-90 sides are 1:1:√2 or √2:√2:2..Area = √2*√2*1/2=1=A
30-60-90 sides are 1:√3:2....Area = 1*√3*1/2=√3/2=1.74/2=0.87=B

Thus A>B
_________________

no.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.