In triangle ABC, angel B=30°, angel C=65° and the bisector of angle A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.
Answers
Given: In ∆ABC, ∠B=35°,∠C=65° and ∠BAX = ∠XAC
To find: Relation between AX, BX and CX in descending order.
In ∆ABC, by the angle sum property, we have
∠A + ∠B + ∠C = 180°
∠A + 35° + 65° = 180°
∠A + 100° = 180°
∴ ∠A = 80°
But ∠BAX = ∠A = 80° = 40°
Now in ∆ABX,
∠B = 35°
∠BAX = 40
And ∠BXA = 180° - 35° - 40°
= 105°
So, in ∆ABX,
∠B is smallest, so the side opposite is smallest, ie AX is smallest side.
∴ AX < BX …(1)
Now consider ∆AXC,
∠CAX =
× ∠A
× 80° = 40°
∠AXC = 180° - 40° - 65°
= 180° - 105° = 75°
Hence, in ∆AXC we have,
∠CAX = 40°, ∠C = 65°, ∠AXC =75°
∴∠CAX is smallest in ∆AXC
So the side opposite to ∠CAX is shortest
Ie CX is shortest
∴ CX <AX …. (2)
From 1 and 2 ,
BX > AX > CX
Answer:
CX < AX < BX
Step-by-step explanation:
Given: In ∆ABC, ∠B=35°,∠C=65° and ∠BAX = ∠XAC
To find: Relation between AX, BX and CX in descending order.
In ∆ABC, by the angle sum property, we have
∠A + ∠B + ∠C = 180°
∠A + 35° + 65° = 180°
∠A + 100° = 180°
∴ ∠A = 80°
But ∠BAX = ∠A
= × 80° = 40°
Now in ∆ABX,
∠B = 35°
∠BAX = 40
And ∠BXA = 180° - 35° - 40°
= 105°
So, in ∆ABX,
∠B is smallest, so the side opposite is smallest, ie AX is smallest side.
∴ AX < BX …(1)
Now consider ∆AXC,
∠CAX = × ∠A
=× 80° = 40°
∠AXC = 180° - 40° - 65°
= 180° - 105° = 75°
Hence, in ∆AXC we have,
∠CAX = 40°, ∠C = 65°, ∠AXC =75°
∴∠CAX is smallest in ∆AXC
So the side opposite to ∠CAX is shortest
Ie CX is shortest
∴ CX <AX …. (2)
From 1 and 2 ,
CX < AX < BX
This is the required ascending order.
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