✯Correct Question:-

In △ ABC, ∠ B = 35°, ∠ C = 65° and the bisector of ∠ BAC meets BC in P. Arrange AP, BP and CP in descending order.

✯Solution:-

Consider △ ABC

By 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

∠ BAP = ½ ∠ A

So we GET

∠ BAP = ½ (80°)

By DIVISION

∠ BAP = 40°

Consider △ ABP

It is given that ∠ B = 35° and

∠ BAP = 40°

By sum property of a triangle

∠ BAP + ∠ BPA + ∠ PBA = 180°

To find ∠ BPA

∠ BPA = 180° – ∠ BAP – ∠ PBA

By substituting values

∠ BPA = 180° – 35° – 40°

By subtraction

∠ BPA = 180° – 75°

∠ BPA = 105°

We know that ∠ B is the smallest angle and the side opposite to it i.e. AP is the smallest side.

So we get AP < BP ….. (1)

Consider △ APC ∠ CAP = ½ ∠ A

So we get

∠ CAP = ½ (80°)

By division

∠ CAP = 40°

By sum property of a triangle

∠ APC + ∠ CAP + ∠ CPA = 180°

To find ∠ APC

∠ APC = 180° – ∠ CAP – ∠ CPA

By substituting values

∠ APC = 180° – 40° – 65°

So we get

∠ APC = 180° – 105°

By subtraction

∠ APC = 75°

So we know that ∠ CAP is the smallest angle and the side opposite to it i.e. CP is the smallest side.

We get CP < AP …… (2)

By considering equation (1) and (2)

BP > AP > CP

Therefore, BP > AP > CP is the descending order.

✯Answer:-

Therefore, BP > AP > CP is the descending order.

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