Question
In $\triangle\text{ABC},\angle\text{B}=35^\circ,\angle\text{C}=65^\circ$ and the bisector of $\angle\text{BAC}$ meets BC in P. Arrange AP, BP and CP in descending order.
✓
Answer
In angle
A + B + C = 180
A + 35 + 65 = 180
A = 180 - 100
A = 80
So $\angle\text{BAP}$ and $\angle\text{CAP}=\frac{80}{2}=40$
we know that side opposite to the greater angle is longer
In $\triangle\text{BAP}$
we know that side opposite to the greater angle is longer
so BP > AP
In $\triangle\text{CAP}$
ACP (65) > CAP (40)
so AP > CP
so BP > AP > CP
A + B + C = 180
A + 35 + 65 = 180
A = 180 - 100
A = 80
So $\angle\text{BAP}$ and $\angle\text{CAP}=\frac{80}{2}=40$
we know that side opposite to the greater angle is longer
In $\triangle\text{BAP}$
we know that side opposite to the greater angle is longer
so BP > AP
In $\triangle\text{CAP}$
ACP (65) > CAP (40)
so AP > CP
so BP > AP > CP
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