darkdevil8z wrote:

Hi everyone, I couldn't confirm my answer whether it is correct or not, can anyone confirm for me?? Or correct me if I'm wrong??

Thanks.

BC = (10√3)/3

BC = 5.77

angle of C = (cos θ = AC/BC)

since radius of the circle with center C is 5

AC = 5

cos θ = 5/5.77

θ = (sin-1) 5/5.77

angle of C ≈ 30

sin θ = AB/BC

sin 30 = AB/5.77

0.5 = AB/5.77

AB = 2.88

Therefore Quantity B is greater.

Both of you are going a very long way.

Method 1 : darkdevil8z method simplified : Only Mental Mathematic required.

Two sides are given and one angle is 90 degree so try diving both the length and see if you recognise the number.

I just divide (10√3)/3 with 5 and remainder is 2/√3.

I don't remember Cos/tan/cosec/cot/sec but I do remember Sin θ i.e (perpendicular / hypotenuse OR P/B)and you too have to.

This is the only requirement for GRE as far as trigonometry application is required.

I remember 1 - 1/2 - 1/√2 - √3/2 - 1 for Sin θ and recall that the above number is familiar. It is reciprocal of (10√3)/3 by 5

Now I know Sin 60 = √3/2 = 5 divided by (10√3)/3 = AC/CB = perpendicular/Hypotenuse

Therefore angle ABC is 60 degree and side in-front will definitely be bigger.

AC is greater This was simpler method but for those who can't recite Sin θ at that moment, go for Pythagoras theorem.

Method 2 : Pythagoras theorem. method simplified : Only Mental Mathematic required.

ABˆ2 = (10/√3) - 5ˆ2 = 100/3 - 25 = 25/3 = Less than 9

AB = Less than 3

AC=5 is greater. To get more simplified solutions.

Start Follow and Hit Kudos / LikeRegards

Shekhar