IIT JEE , AIEEE Forum - Mathematics , Trignometry

subhashree

1)in a triangle ABC, if the angular bisector AD and the median BE are perpendicular then BE/AD = ?
OPTIONS: a) 3/2 tanA/2 b)3 cotA/2 c)2/3 cotA/2 d)5/8 tanA/2

2)In triangle ABC, AD is the median. AE , AF are the medians in the triangles ABD , ADC respectively. If AD = 1 , AE = 2 , AF = 3 , then a² = ?

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subhashree

respected sir,
i am very sorry for the delay in response.actually i don't have net at home so i've to go out for and it is just not possible for me do that daily.

sir want the first question to be answered.

kghedriu

A1) in triangle ABD, we have tanA/2= BD/AD = 2/3*BE/AD (since BE is a median, D divides in ratio 2:1 frm B)

so, BE/AD= 3/2*tanA/2 (option (a))

rate me if u hav understood...

rajat

hey goutham,just check,you have taken angle ADB to be 90.that's not possible.

i suggest you read the question once more.also, i think D should lie on BC and E lies on AC.i am not getting the answer .

vinu

Hi subhashree,

A1)

ABE is isosceles....So, AB = AE ;{ i.e; c= b/2} ;

In

ABD , (AD /Sin B ) = (BD /Sin A/2); { BD= ca /(b+c) } ;

u get AD = (2/3)b*Cos (A/2) ..........i) ;

BE =2*AE*Sin (A/2) =bSin (A/2) ;

Hence, BE/AD = (3/2) Tan (A/2) .

Option a) is the ans.

A2) a² = 88.

rajat

hey vinu,even i got the same answer using different method but how did u get

BD=ac/b+c ??

vinu

Hi rajat,

An angular bisector of an angle divides the opp side in the ratio of other 2 sides....

i.e ; D divides BC in the ratio c:b ;

So, BD = ca / (b+c).

and..is the diff method of urs simpler than mine?

if so, will u plz post ur solu....

rajat

you mean BD/DC = c/b

coooool i did not know this

well, once you know this formula, your method becomes shorter than mine any day. by the way( although my solution has merely two lines more than yours)

where did you come across this concept ?? some book or coaching class shortcut ??

plzz. let me know more such small-small concepts ( not formulae plzzz. ,only concepts which we tend to forget )

and how did u do the 2nd one ???

subhashree

hay venu, how did u arrive at AD=2/3bcosA/2

vinu

Hi subhashree,

Just use the relation given above , and use c=b/2 , Sin B = bSin A/a ...

u'll get AD.

And rajat, it's there in every damn book. I think u just forgot that 1.

amay

arey rajat, read the qn, it was given that median and angular bisector are perpendicular.

waise, vinu, tumhara soln bhi achcha tha....

kghedrui, tell me shortcut how to call u, its gettin tougher to type u name...

puneet

hiiii

Well let me answer the first part for u .

Let P be the point where AD and BE meet

clearly since AD is perpendicluar to BE and AD is perpendicular bisector so,

triangle ABP and triangle APE are congurent and hence BP = PE

Also , c = b/2 ( AB = AE )

and, thus BE = 2BP = 2.c.sin(A/2)

Also we know length of angle bisector AD = [2.b.c/(b+c).]cos(A/2) [this is a standard formula]

Now since c = b/2

so, AD = 2/3.b.cos(A/2)

SO, BE/AD = (b.sinA/2) / (2/3.b.cos(A/2) )

= 3/2.tan(A/2)

Hence answer is (A) ..

cheers

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