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Question
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
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Solution
It is given that
∠B = 35°
∠C = 65°
AP is the bisector of ∠CAB
We have to arrangeAP, BPand CPin descending order.
In ΔACP we have
∠ACP = 65°
∠CAP = 40°(As AP is the bisector of ∠CAB
So AP > CP (Sides in front or greater angle will be greater) ........(1)
In ΔABP we have
∠BAP = 40°(As AP is the bisector of ∠CAB)
Since,
∠BAP >∠ABP
So BP > AP ..........(2)
Hence
From (1) & (2) we have
BP > AP > CP
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