Q. 383.8( 6 Votes )
In ∆ABC, ∠B=35°,∠C=65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.
Answer :
Given: In ∆ABC, ∠B=35°,∠C=65° and ∠BAX = ∠XAC
To find: Relation between AX, BX and CX in descending order.
In ∆ABC, by the angle sum property, we have
∠A + ∠B + ∠C = 180°
∠A + 35° + 65° = 180°
∠A + 100° = 180°
∴ ∠A = 80°
But ∠BAX = 
∠A
= 
× 80° = 40°
Now in ∆ABX,
∠B = 35°
∠BAX = 40
And ∠BXA = 180° - 35° - 40°
= 105°
So, in ∆ABX,
∠B is smallest, so the side opposite is smallest, ie AX is smallest side.
∴ AX < BX …(1)
Now consider ∆AXC,
∠CAX = 
× ∠A
=
× 80° = 40°
∠AXC = 180° - 40° - 65°
= 180° - 105° = 75°
Hence, in ∆AXC we have,
∠CAX = 40°, ∠C = 65°, ∠AXC =75°
∴∠CAX is smallest in ∆AXC
So the side opposite to ∠CAX is shortest
Ie CX is shortest
∴ CX <AX …. (2)
From 1 and 2 ,
BX > AX > CX
This is required descending order
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