Question

help i will mark brillent

Answer 1

In ΔABC, by angle sum property we have 
2x + 2y + ∠A = 180° 
⇒ x + y + (∠A/2) = 90°
 ⇒ x + y = 90° –  (∠A/2)  à (1)
 In ΔBOC, we have
  x + y + ∠BOC = 180° 90° – (∠A/2) + ∠BOC = 180° [From (1)] 
∠BOC = 180° – 90° + (∠A/2) ∠BOC = 90° + (∠A/2)......



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In a triangle, the sum of total angle =180°

Thus in ∆ABC, [math]\angle A+\angle B+\angle C=180°[/math]

[math]\Rightarrow 60°+\angle B+\angle C=180°[/math]

[math]\Rightarrow\angle B+\angle C=180°- 60°=120°[/math]

[math]\Rightarrow\frac{\angle B+\angle C}{2}=\frac{120°}{2}[/math]

[math]\Rightarrow\frac{\angle B}{2}[/math][math]+\frac{\angle C}{2}=60°[/math]

[math]\Rightarrow\angle OBC+\angle OCB=60°[/math]

Thus, in ∆OBC,[math]\angle OBC+\angle OCB+\angle BOC=180°[/math]

[math]\Rightarrow\angle BOC=180°-60°=120°[/math]

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