Question

In the given fig., ABC, ZB = 36°, ZC = 64° and bisector of ZBAC meets BC in X. Arrange AX, BX and CX in ascending order.​

Answer 1

Answer:

Consider △ ABC By sum property of a triangle ∠ A + ∠ B + ∠ C = 180o To find ∠ A ∠ A = 180o – ∠ B – ∠ C By substituting the values ∠ A = 180o – 35o – 65o By subtraction ∠ A = 180o – 100o ∠ A = 80o We know that ∠ BAX = ½ ∠ A So we get ∠ BAX = ½ (80o) By division ∠ BAX = 40o Consider △ ABX It is given that ∠ B = 35o and ∠ BAX = 40o By sum property of a triangle ∠ BAX + ∠ BXA + ∠ XBA = 180o To find ∠ BXA ∠ BXA = 180o – ∠ BAX – ∠ XBA By substituting values ∠ BXA = 180o – 35o – 40o By subtraction ∠ BXA = 180o – 75o ∠ BXA = 105o We know that ∠ B is the smallest angle and the side opposite to it i.e. AX is the smallest side. So we get AX < BX ….. (1) Consider △ AXC ∠ CAX = ½ ∠ A So we get ∠ CAX = ½ (80o) By division ∠ CAX = 40o By sum property of a triangle ∠ AXC + ∠ CAX + ∠ CXA = 180o To find ∠ AXC ∠ AXC = 180o – ∠ CAX – ∠ CXA By substituting values ∠ AXC = 180o – 40o – 65o So we get ∠ AXC = 180o – 105o By subtraction ∠ AXC = 75o So we know that ∠ CAX is the smallest angle and the side opposite to it i.e. CX is the smallest side. We get CX < AX …… (2) By considering equation (1) and (2) BX > AX > CX Therefore, BX > AX > CX is the descending order.

Explanation:

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