Questions number 25 from attached image. Also the question number 24

Dear student 

Image is not clear , please upload the fresh imge , 

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Good luck 

 

hello student, ignore vikas.... the image is clear. he doesnt download it though. and pratisksh kindly ask only 1 ques in one thread.

now, lets solve 25 th ques.

you can draw the figure and let BD= n so that AB= n+3 and let AC= m. let angle ABC= 6x so that BAC= 90 – 3x (due to properties of isosceles triangle).

now, sine rule in ABD will give you

(n+3)cos3x= ncosx ….......(1) which you can simplify by expanding cos3x.

also, m= 2(n+3)*cos(90 – 3x)= 2(n+3)sin3x............(2)

from (1) and (2) eliminate x to obtain 

27p^2 – m^2p – m^2= 0 where p= n+2 and since n is an integer greater than or equal to 1, p will be greater than or equal to 3. note that eliminating x from (1) and (2) will be a bit lengthy. 

now, by quad formula you ll be able to express p in terms of m and finally note that the discriminant D (after simplification) will be m^2 – 108. now, since p is also an integer, D would have to be square of an integer.

so, m^2 – 108= k^2

or (m – k)(m+k)= 108

now, we can easily find factor pairs of 108 as (1, 108); (2, 54); (3, 36); (4, 27); (6, 18); (9, 12).

so you ll have to check each of these pairs, and finally only 2 of them will work ((2, 54) and (6, 18)) which will give you m= 6, 26.

corresponding to these values of m, the values of p will be 2 and 26. since p has to b greater than equal to 3, hence 2 is rejected. 

so, m= 26

i know this is an insanely hard and specially lengthy problem...... but i hope u can understand .