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In DeltaABC, angleB=35^(@), angleC=65^(@...

In `DeltaABC, angleB=35^(@), angleC=65^(@)` and the bisector of `angleBAC` meets BC in X. Arrange AX, BX and CX in descending order.

Text Solution

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`angleA+35^(@)+65^(@)=180^(@)rArr angleA=80^(@).`
`therefore" "angleBAX=angleCAX=40^(@), angleABX=35^(@), angleACX=65^(@)`.
`angleAXB=180^(@)-(40^(@)+35^(@))=105^(@).`
`angleBAX gt angleABX rArr BX gt AX.`
`angleACX gt angleCAX rArr AX gtCX.`
`therefore" "BXgtAXgtCX.`
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