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Since the angles are supplementary therefore the sum of their angles will be 180 degree.
As we know that the angles are supplementary so sum of angles will be 180 degree.
Let us assume that the ratio factor is r.
According to question,
Angles are supplementary and have a ratio of 1:4.
r + 4r = 180
⇒ 5r = 180
⇒ r = 180/5
⇒ r = 36
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complement of 30°20′ = 90° – ( 30°20′ ) = 90° – ( 30° + 20′ )
= (89° – 30°) + (1° – 20′)
complement of 30°20′ = 90° – ( 30°20′ ) = 90° – ( 30° + 20′ )
= (89° – 30°) + (1° – 20′)
= 59° + 60′ – 20′ [ ∴ 1° = 60°′]
= 59° + 40′ = 59°40′.
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Through O, draw a line l parallel to both AB and CD. Then
∠1 = 45° (alt. ∠S)
and ∠2 = 30° (alt. ∠S)
∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75°
Through O, draw a line l parallel to both AB and CD. Then
∠1 = 45° (alt. ∠S)
and ∠2 = 30° (alt. ∠S)
∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75°
So, X = 360° – ∠BOC = 360° – 75° = 285°
Hence X = 285°.
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Let the measured of the required angle be P degree.
Then, its supplement = 180 – P
Now use the formula,
Angle | = | 1 | ( its supplement ) |
3 |
Let the measured of the required angle be P degree.
Then, its supplement = 180 – P
Now use the formula,
Angle | = | 1 | ( its supplement ) |
3 |
P | = | ( 180 - P ) |
3 |
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AB || CD and t transversal intersects them at E and F
∠BEF + ∠EFD = 180° (co-interior angles)
⇒ | 1 | ∠BEF | + | 1 | ∠EFD | = 180/2 |
2 | 2 |
AB || CD and t transversal intersects them at E and F
∠BEF + ∠EFD = 180° (co-interior angles)
⇒ | 1 | ∠BEF | + | 1 | ∠EFD | = 180/2 |
2 | 2 |
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