@sonukg4india Let the 36 be x×(72/x), and the 20 be (40/y)×y.
Then (y-72/x)(x-40/y)=32.
xy + 2880/xy = 144
A²-144A+2880=0=(A-120)(A-24)
A = 36+20+16+?, so A=120 and ?=48.
@hope_dies_again@Ginger_Tucci@CJ_7171 I assume you are referring here to the Palestine Action activists, who have not in fact been found guilty of any offence. No doubt you know more about all that than the jury did, though.
@ybgoi Mass points for this configuration are A2, B4, C1, giving D6 and E3. Thus DP:PC=1:6, and [ADP]=⅔×⅙×[ABC]=⅑[ABC].
Now let vector BC=p and BA=q.
So DC is p-⅓q, PA=-⅐(p-⅓q)+⅔q.
So directions are 3p-q, -p+5q.
The dot is -3p.p+16p.q-5q.q=0, so perpendicular.
@ai_meh_s@Asadthegeek@BiancaDante27@viterick@il1123@ohmohm I don't mean to be rude, and I appreciate the compliment inherent, but I now have about sixty notifications for these threads. Please don't tag me in problems (unless I wrote them).
@sonukg4india Masses A(1), B(1), C(2) give F at centre. So CD=FD and AF=3EF. So [DAF]=[CAF]=3[CEF], which means the answer is ⅓. The equal angles are irrelevant.