Well in that case, the length of hypotnuse should be given.

T₄ = 0

And the angle be @, and hypotnuse be x.

Then T₃ = T₅ = T (say) ...... as same string.

Then we get,

T₁ = T.sin @ = 10g

T₂ = T.cos@

 

Hence, T = 9g/x

and T₂ = ((8100 - x²))g / 9

 

This will do .........

 

But now also I think that acc to the figure, 37 and 53 are the angles ........

I think, for minimum difference the pt P should be the point of intersection should of the line AB and y=x.

Hence P == (2.6 , 2.6)

Hence, PA - PB = (68 - 153) / 5.

 

I know, the ans is weird but this is all i could think of.

The whole system is in equilibrium.

Hence, the equations will be .....

(1) T₁=mg

(2) At the lower point,

     T₃.cos 37 = T₁

     T₃.sin 37 = T₂

(3) At upper point,

     T₅.cos 53 = T₃.cos 37

     T₅.sin 53 = T₄ + T₃.sin 37

 

and we know that tan 37 = 4 / 3

 

Hence,

T₁ = 10g

T₂ = 7.5g

T₃ = 12.5g

T₄ =  5.84g

T₅ = 16.67g

 

Hope you get it .....

This book gives us systematic solutions of the JEE papers (both Screening and Mains). It even helps students to know the way of application of their knowledge in getting the solutions of the questions asked. It makes acquainted with the styles, types and extent of difficulty by going through lucid and elaborate solutions.

 

Moreover, I found that this book contains less errors in the solutions than other books available in the market.

Great work yaar ........

Tu to mera calculus sudhravaega ........

The vertices of the triangle are A(-8,5), B(-15,-19) & C(1,-7).

Hence AB = 25, BC = 20 & CA = 15.

 

Now theres a prop of angle bisector that it divides the opposites side in the ratio of its arms. Hence AD divides BC in the ratio 5:3.

 

Hence, D(-5,-23/2).

Now, we can write the equation of AD in two point form.

It is (-13/3)x + 2y - (5/3) = 0

 

Hence a = (-13/3) and c = (-5/3)

 

I think its Self-confidence , Desire to get into da IITs, Smartwork (Smart studies) and Creativity .......