Question

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Answer 1

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Given :

• ∆ABO
• External angle ∠POQ=40°
• ∠1:∠2=4:3

To find :

• ∠1-∠2

Solution :

∠1:∠2=4:3

Let ∠1 be 4x

and ∠2 be 3x

Line AQ and PB interest each other

∠POQ=∠AOB. [vertically opposite angles]

∠AOB=40°

Sum of all angles of triangle is 180°

In ∆ ABO

:⟶∠ABO+∠BOA+∠BAO = 180°

:⟶∠1+40°+∠2 = 180°

:⟶4x+40°+3x=180°

:⟶7x=180°-40°

:⟶7x=140°

:⟶x=140°/7

:⟶x=20°

• ∠1 = 4x =4×20°=80°
• ∠2 =3x=3×20°=60°

=∠1-∠2

=4x-3x

=80°-60°

=20°

Answer 2

Answer:

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Explanation: