Question

Linges and angles
plz answer ​

Answer 1

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In ΔABC, we have

∠A + ∠B + ∠C = 180° (sum of the angles of a triangle)

⇒ 1/2∠A + 1/2∠B + 1/2∠C = 90°

⇒ 1/2 ∠A + ∠x + ∠y = 90° ….(i)

Again in ΔOBC

∠OBC + ∠OCB + ∠BOC = 180° (by angle sum property of a triangle)

⇒ ∠x + ∠y + ∠BOC = 180° …(ii)

Substituting ∠x + ∠y = 90° – 1/2 ∠A from (i) in eqn (ii),

we get 90° – 1/2∠A + ∠BOC = 180°

⇒ ∠BOC = 180° – 90° + 1212 ∠A

⇒ ∠BOC = 90° + 1/2<A

Answer 2

Given :-

In ΔABC , The bisector of AngleB and Angle C intersect each other at point O.

To prove :-

Angle BOC = 90° + 1/2

Proof :-

In ΔBOC , Let's consider Angle COB = Angle 1 and Angle OCB = Angle 2

Now, By using Angle Sum Property,

Δ1 + ΔBOC + Δ2 = 180° ( 1 )

Now, In ΔABC

By using Angle sum property,

ΔA + ΔB + ΔC = 180°

ΔA + 2Δ1 + 2Δ2 = 180°

[ Here, we have wrote 2Δ1 and Δ2Δ2 because the bisector divides it into two parts ]

ΔA + 2( Δ1 + Δ2 ) = 180°

ΔA/2 + Δ1 + Δ2 = 180° /2

ΔA/2 + Δ1 + Δ2 = 90°

Δ1 + Δ2 = 90° - ΔA/2

From eqn( 1 )

90° - ΔA/2 + ΔBOC = 180°

-ΔA/2 + ΔBOC = 180° - 90°

ΔBOC = 90° + ΔA/2

Hence proved $.$