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JEE Advanced · Mathematics · 17. Properties of Triangles

In a XYZ let x,y,z,  be the lengths of sides opposite to the angles,  X, Y, Z, respectively and 2s=x+y+z. If s-x4=s-y3=s-z2 and area of incircle of the triangle XYZ is 8π3, then

  1. A Area of the triangle XYZ is 66
  2. B The radius of circumcircle of the triangle XYZ is 3566
  3. C     sinX2sinY2sinZ2=435
  4. D sin2X+Y2=35
Verified Solution

Answer & Solution

Correct Answer

(A) Area of the triangle XYZ is 66

Step-by-step Solution

Detailed explanation


Let s-x4=s-y3=s-z2=k (K0)
s-x=4k
s-y=3k
s-z=2k
Adding this, we get
3s-x+y+z= 9k
s=9k
x=5k, y=6k, z=7k

πr2=8π3
πΔs2=8π3
= 83 .s
ss-x s-ys-z= 83. s
9k. 4k. 3k. 2k= 83 9k
24.9 k2= 83 . 9k 3×2 6 K=223 9
K 22.966 3=3232=1
k=1
x=5, y=6, z=7
Δ= 83 . 9k= 83 .9=66
R = circumradius =xyz4Δ=5. 6. 74..66=3546
Using formula sinX2sinY2sinZ2=r4R=ΔsxyzΔ=Δ2s.xyz=36.69.5.6.7=435
cosZ= x2+y2-z22xy=25+36-492.5.6=15
sin2X+Y2 = cos2Z2=1+cosZ2=1+152=35
From JEE Advanced
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