Given a right triangle ABC right angled at C and whose legs are given `1+4log_(p^2)(2p),1+2^(log_2(log_2(p))` and hypotenuse is given to be `1+log_2(4p)`. The area of `trianleABC` and circle circumscribing it are `Delta_1` and`Delta_2` respectively.

Given a right triangle ABC right angled at C and whose legs are given 1+4log_(p^(2))(2p), 1+2^(log_(2)(log_(2)p)) and hypotenuse is given to be 1+log_(2)(4p) . The area of DeltaABC and circle circumscribing it are Delta_(1) and Delta_(2) respectively, then Q. The value of sin ((pi(25p^(2)Delta_(1)+2))/(6))=

Given a right triangle ABC right angled at C and whose legs are given 1+4log_(p^(2))(2p), 1+2^(log_(2)(log_(2)p)) and hypotenuse is given to be 1+log_(2)(4p) . The are of DeltaABC and circle circumscribing it are Delta_(1) and Delta_(2) respectively, then Q. Delta_(1)+(4Delta_(2))/(pi) is equal to :

Given a right triangle ABC right angled at C and whose legs are given 1+4log_(p^(2))(2p), 1+2^(log_(2)(log_(2)p)) and hypotenuse is given to be 1+log_(2)(4p) . The are of DeltaABC and circle circumscribing it are Delta_(1) and Delta_(2) respectively, then Q. Delta_(1)+(4Delta_(2))/(pi) is equal to :

Solve : log_8 (p^2 - p) - log_8 (p-1) = 2

Solve log_(8)(P^(2)-p)-log_(8)(p-1)=2

In the given figure ABC is a right triangle and right angled at B such that /_BCA = 2/_BAC. Show that hypotenuse AC = 2BC.

In the given figure ABC is a right triangle and right angled at B such that /_BCA = 2/_BAC. Show that hypotenuse AC = 2BC.

In the given figure ABC is a right triangle and right angled at B such that /_BCA = 2/_BAC. Show that hypotenuse AC = 2BC.

In the given figure ABC is a right triangle and right angled at B such that /_BCA = 2/_BAC. Show that hypotenuse AC = 2BC.