IIT JEE , AIEEE Forum - Mathematics , Trignometry

elastiboysai

Here's a gud qn.

not tough...

If the sides of a triangle are in A.P and its greatest angle exceeds the least angle by

find the ratio of the sides in the form

1+x:1:1-x

hsbhatt

I dont know if this is the smartest way to go about it.

Let the sides be a-d,a, a+d.

Then the ratio in the required format is 1-a/d:a:1+a/d.

Hence, our task is to find a/d

a-d = k sinA; a = k sinB; a+d = k sinc

Since, sinA, sinB, sinC are also in AP

2sinB = sinA+sinC

which gives 2cos(A+C/2) = cos

/2

Now a/d = sinA+sinC/(sinC-sinA) = tan(A+C/2) cot

/2

Since cos(A+C/2) = 0.5*cos

/2, tan(A+C/2) =

(4-cos²

/2)/cos(

/2)

Hence a/d =

(4-cos²

/2)/cos(

/2) * cot

/2 =

(4-cos²

/2) /sin(

/2)

elastiboysai

edited

haan sir now that calc mist. is rectified

btw

plz give d ans exclusively in terms of cos

if u can..

Radon222

I am getting something weird always an equilateral triangle.

let the angles be

$\theta-\alpha/2 ,\theta,\theta+\alpha/2 \\ \\ \theta-\alpha/2 + \theta + \theta+\alpha/2=180^o\\ \\ \theta=60 \\ \\ \text{now using cosine rule} \\ \\ cos60=\frac{(a+d)^2+(a-d)^2-a^2}{2(a^2-d^2)}\\ \\ a^2-d^2=2a^2+2d^2-a^2\\ \\ d=0$

why is this happening???

nadeemoidu

@radon

first of all its not the angles that are in AP , it's only the sides.

Also , in the final step, u get
$3d^2 = 1 \\ \\ d= \pm \frac{1}{\sqrt{3}}$

Radon222

Oh ! damn .....this is the reason why we should not solve problems in late night

raulrag009

elastiboysai

nope not compl rt.

btw

culd u post d ans in terms of cos

alone??

elastiboysai

Its high time i posted my soln

this 1 s a standard 1

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