IIT JEE , AIEEE Forum - Mathematics , Trignometry

nigitha_17

in triangle ABC if the incenter is middle point of the median AD 'then the value of cosA is...

nigitha_17

plz see this

nigitha_17

would any one plz bother to answere it

joyfrancis

---refer to next post----

puneet

hiii

well, the method followed by joy is correct ..

there are various ways of going about it .. bu tu will reach thr only ..

I hope this is clear ..

cheers

bunnyaneja

Sorry, for trying to prove an iitian and a scorching goiitian wrong.

Mr.Joy check out the statements.

You are trying to say that

1-1/2 [2 sin(A/2)]^2 = [cos(A/2)] ^2

How the hell is that possible???????!!!!!!!!!!!!!!

Everyone knows that 1-(sin x)^2 = (cos x)^2

But, 1- 2 (sin x)^2 is not = (cos x)^2, rather it is equal to cos 2x.

ankurgupta91

hey bunnyaneja is correct
it shd be like this
1-1/2 [2 sin(A/2)]^2 = cosA
which doesnt give nything
nd moreover cosA =1 which means
A =0 which is nt possible at all...........

joyfrancis

yup!silly mistake....I will post the correct solution very soon.

joyfrancis

Angle ADB = 180 - (A/2+B)

so ADC=A/2 + B

since sin of both these angles is same we have AB=AC

now ADB is congruent to ADC (SAS rule)

andles ADB and ADC are each equal to 90.

also, angle B = angle C = 90-A/2

.:2cosA/2 = a/b

cosine rule: cosA = b^2+c^2-a^2 / 2bc

cosA = 1 - 1/2 (a/b)^2

= 1 - 2 cos^2A/2

= - cosA

2cosA = 0

cosA=0

A=90

this should be it..

thanks

joyfrancis

nigitha_17

is this the correct answer---cosA=0

joyfrancis

see this for clarity.

nigitha_17

no the ans is1/4.

joyfrancis

please solve this question.. i'm unable to get the right answer!

Ask & Discuss Questions with Community & Experts