Consider
△ABCBy sum property of a triangle
∠A+∠B+∠C=180∘
To find ∠A
∠A=180∘−∠B−∠C
By substituting the values
∠A=180∘−35∘−65∘
By subtraction
∠A=180∘−100∘
∠A=80∘
We know that
∠BAX=12∠A
So we get
∠BAX=12(80∘)
By division
∠BAX=40∘
Consider △ABX
It is given that ∠B=35∘ and ∠BAX=40∘
By sum property of a triangle
∠BAX+∠BXA+∠XBA=180∘
To find ∠BXA
∠BXA=180∘−∠BAX−∠XBA
By substituting values
∠BXA=180∘−35∘−40∘
By subtraction
∠BXA=180∘−75∘
∠BXA=105∘
We know that ∠B is the smallest angle and the side opposite to it i.e. AX is the smallest side.
So we get AX<BX..(1)
Consider △AXC
∠CAX=12∠A
So we get
∠CAX=12(80∘)
By division
∠CAX=40∘
By sum property of a triangle
∠AXC+∠CAX+∠CXA=180∘
To find ∠AXC
∠AXC=180∘−∠CAX−∠CXA
By substituting values
∠AXC=180∘−40∘−65∘
So we get
∠AXC=180∘−105∘
By subtraction
∠AXC=75∘
So we know that ∠CAX is the smallest angle and the side opposite to it i.e. CX is the smallest side.
We get
CX<AX(2)
By considering equation (1) and (2)
BX>AX>CX
Therefore, BX>AX>CX is the descending order.