19. In the given figure, BI is the bisector ofABC and CI is the bisector of ACB, Find BIC.
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please insert your figure dear it is not sufficient to gave solution
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Step-by-step explanation:
In ∆ ABC,
BI is the bisector of ∠ABC and CI is the bisector of ∠ACB.
∵ AB = AC
∴ ∠B = ∠C
(Angles opposite to equal sides)
But ∠A = 40°
and ∠A + ∠B + ∠C = 180°
(Angles of a triangle)
⇒ 40° + ∠B + ∠B = 180°
⇒ 40° + 2∠B = 180°
⇒ 2∠B = 180° - 40° = 140°
⇒ ∠B = 140°/2 = 70°
∴ ∠ABC = ∠ACB = 70°
But BI and Cl are the bisectors of ∠ABC and ∠ACB respectively.
∠IBC = 1/2 ∠ABC = 1/2 (70°) = 3
and ∠ICB = 1/2 ∠ACB = 1/2 × 70°= 35
Now in ∆ IBC,
⇒ ∠BIC + ∠IBC + ∠ICB = 180°
(Angles of a triangle)
⇒ ∠BIC + 35° + 35° = 180°
⇒ ∠BIC = 180° - 70° = 110°
Hence ∠BIC = 110°
Hope it helps you
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