Heya
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Heya!
Here is yr answer......
➡ Show that the bisectors of the base angles of a triangle can never enclose a right triangle.
Sol :
Let us take a Δ ABC
Bx is the bisector of angle B and Cy is the bisector of angle C
Here, p is the point formed by the bisectors!
Since, p is greater than 90°.
Therefore, The bisectors of base angles of a triangle can never enclose a right triangle.
HENCE PROVED!
Hope it hlpz..
Here is yr answer......
➡ Show that the bisectors of the base angles of a triangle can never enclose a right triangle.
Sol :
Let us take a Δ ABC
Bx is the bisector of angle B and Cy is the bisector of angle C
Here, p is the point formed by the bisectors!
Since, p is greater than 90°.
Therefore, The bisectors of base angles of a triangle can never enclose a right triangle.
HENCE PROVED!
Hope it hlpz..
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tejasri2:
great answer
Answered by
5
★Heya !!★
Here's your solution !!!
_______________________________
☛In ∆ABC, BP and CP are bisectors of angles B and C respectively.
Hence,
∠A + ∠B + ∠C = 180°
∠A + 2∠1 + 2∠2 = 180°
2(∠1 + ∠2) = 180° - ∠A
(∠1 + ∠2) = 90° - (∠A/2) ··········(1)
In ∆PBC, ∠P + ∠1 + ∠2 = 180°
∠P + [90° - (∠A/2) = 180° [From (1) ]
∠P = 180° - [90°-(∠A/2)
[90° + (∠A/2)
Hence, angle P is always greater than 90°.
thus PCB can never be a right angle
triangle.
_____________________________
GLAD HELP YOU,
It helps you,
thank you☻
@vaibhav246⚑
Here's your solution !!!
_______________________________
☛In ∆ABC, BP and CP are bisectors of angles B and C respectively.
Hence,
∠A + ∠B + ∠C = 180°
∠A + 2∠1 + 2∠2 = 180°
2(∠1 + ∠2) = 180° - ∠A
(∠1 + ∠2) = 90° - (∠A/2) ··········(1)
In ∆PBC, ∠P + ∠1 + ∠2 = 180°
∠P + [90° - (∠A/2) = 180° [From (1) ]
∠P = 180° - [90°-(∠A/2)
[90° + (∠A/2)
Hence, angle P is always greater than 90°.
thus PCB can never be a right angle
triangle.
_____________________________
GLAD HELP YOU,
It helps you,
thank you☻
@vaibhav246⚑
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