1Ans 1 ) 32 ??
in a ΔABC , radii of escribed circles are r1, r2 , r3 and radius of incircle is r.
1) A(z1) ; B(z2) ; C(z3) are the vertices of a ΔABC in argand plane and │z1-α│=│z2-α│=│z3-α│=8 then r1+r2+r3-r is......
2) in a ΔABC , the radius of escribed circles opposite to the vertices B & C are r1 & r2 given by 6, 18 respectively. then the length of altitude of ΔABC through A is......
3) in a ΔABC, tan(A/2) = 5/6 ; tan(C/2) = 20/37 then....
a) a, b, c are in A.P
b) a , c, b are in A.P
c) a, b, c are in G.P
d) a, b, c are in H.P
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9 Answers
1aveek...ur ans for the 3rd ques z rong yar....! itz option b....
in a triangle A+B+C =1800
A/2 +B/2 +C/2 = 900
so...A/2 +B/2 =900- C/2
applyin tan on both sides...
tan(A/2) and tan(C/2) values are given...
finally...we get tan(B/2) =2/5
implies...tan(A/2).tan(B/2) =1/3
using the formulae of tan(A/2) and tan(b/2) interms of s and the sides of the triangle.... we get a+b =2c
so, a,c,b are in A.P...
11) consider α as a point in the argand plane...
z1-α│=│z2-α│=│z3-α│=8 represents the point S equidistant from the vertices A, B and C
SA=SB=SC=8
so, S is the circumcentre of the triangle with a circumradius of R=8 units
we have the formula r1+r2+r3-r = 4R
so.. the ans is 32.
2) 1h1+1h2+1h3=1r
1h2+1h3-1h1=1r1
1h1+1h3-1h2=1r2
1h1+1h2-1h3=1r3
where h1, h2 , h3 are the lengths of the altitudes from the vertices A,B,C respectively.
1r2+1r3 = 2h1
putting the values of r2 and r3 we get h1 = 9
3) itz in d previous post,....!
1hmmm misread the options in question 3.
agar process malum hai to pucha hi kyu ????
62its not always about asking doubts i guess.. sometimes it is simply about sharing something good at times..
1@aveek.... i gt d idea of it on d nxt day wen i ws chekin out 4 sme formula in trigo....! lolzz....:)