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RD Chapter 11 Co ordinate Geometry Ex 11.1 Solutions

Question - 11 : - In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.

Answer - 11 : -

Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB
 
To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° +  ∠A
But ∠BOC = 120°
∴ 90°+   ∠A = 120°
∴ 12 ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C =  = 60°
Hence ∠A = ∠B = ∠C = 60°

Question - 12 : - If each angle of a triangle is less than the sum ofthe other two, show that the triangle is acute angled.

Answer - 12 : -

In a ∆ABC,
Let ∠A < ∠B + ∠C
 
⇒∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)
Similarly, we can prove that
∠B < 90° and ∠C < 90°
∴ Each angle of the triangle are acute angle.

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