Question

In the given figure, AB || CD, ∠ABE = 120°, ∠ECD = 100° and ∠BEC = x°. Find the value of x​

Answer 1

$\bf\underline{\underline{\pink{Question:-}}}$

★In the given figure, AB || CD, ∠ABE = 120°, ∠ECD = 100° and ∠BEC = x°. Find the value of x

$\bf\underline{\underline{\blue{Given:-}}}$

★In the given figure, AB || CD, ∠ABE = 120°, ∠ECD = 100°

$\bf\underline{\underline{\green{ToFind:-}}}$

★The value of x

$\bf\underline{\underline{\red{Construction:-}}}$

★Through E, draw FEG || AB || CD.

$\bf\underline{\underline{\orange{Solution:-}}}$

Now, FE || AB And BE is the transversal.

∴ ∠BEF + ∠ABE = 180°

==> ∠BEF + 120° = 180° [co-interior angle]

==> ∠BEF = (180° – 120°)

==> ∠BEF = 60°

Again, CD || EG and CE is the transversal.

∴ ∠CEG + ∠ECD = 180° [co-interior angle]

==> ∠CEG + 100° = 180°

==> ∠CEG = (180° – 100°)

==> ∠CEG = 80°

Now, FEG is a straight line.

∴ ∠BEF + ∠BEC + ∠CEG = 180°

==> 60° + x° + 80° = 180°

==> x + 140° = 180°

==> x = (180° – 140°)

==> x = 40°

Hence, the value of x is 40°

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Answer 2

Given:

AB || CD

∠ABE = 120°

∠ECD = 100°

∠BEC = x°

Find: value of x​

Through E, draw  FEG || AB || CD. Now, FE ∣∣ A B FE∣∣AB and BE is the transversal.

• ∠ B E F + ∠ A B E = 180°  [co-interior ∠ s ∠s]

• ∠ B E F + 120°  = 180°

• ∠BEF+120°=180°

• ∠ B E F = ( 180° − 120° ) = 60°

∠BEF=(180°-120°)=60°. Again C D ∣ ∣ E G CD∣∣EG and CE is the transversal.

• ∠ C E G + ∠ E C D = 180°
• ∠CEG+∠ECD=180°  [co-interior ∠ s ∠s]  
• ∠ C E G + 100°  = 180°
• ∠CEG+100∘=180°
• ∠ C E G = ( 180° − 100° ) = 80°

Now, FEG is a straight line.  

• ∠ B E F + ∠ B E C + ∠ C E G = 180°
• ∠BEF+∠BEC+∠CEG=180°
• 60° + x° + 80° = 180°
• x + 140° = 180°
• x = 180° - 140°
• x = 40°

Hence the value of x is 40°