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Ratio between angles and sides in a right angle traingle. [#permalink]
02 May 2012, 08:32
Is there a general formula or an idea to calculate the ratio between the angles when sides are given and/or vice-versa? Can this be applied to non-right angle triangles too?
Re: Ratio between angles and sides in a right angle traingle. [#permalink]
02 May 2012, 09:23
@Jay: There is ,in the case of a right angle triangle. One has to do some manipulations over the base, perpendicular and hypotenuse of the angles. I have very little memory of what has to be done exactly.
Re: Ratio between angles and sides in a right angle traingle. [#permalink]
21 May 2012, 14:26
No relation.
A 30-60-90 triangle with sides of 3-4-5 has ratios of 1:2:3 (angles) and 3:4:5 (sides). It could also have the same angle ratios but have side ratios as 5:12:13.
Re: Ratio between angles and sides in a right angle traingle. [#permalink]
27 May 2012, 05:24
Hi,
Here is the general formula (applicable to any triangle), \(a/sinA = b/sinB = c/sinC\) (out of scope of GMAT)
And as per GMAT, for a right angles triangle only two combinations of angles are useful, i.e, 30,60,90 where ratio of sides is 1:\sqrt{3}:2 and 45,45,90 -> ratio of sides - 1:1:\sqrt{2}
Regards,
Last edited by cyberjadugar on 27 May 2012, 08:39, edited 1 time in total.
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