Question

in figure L1 // L2, the value of x is?? ​

Answer 1

Solution :-

In ∆ABC we have,

→ 45° + 35° + ∠CAB = 180° { By angle sum property. }

→ 80° + ∠CAB = 180°

→ ∠CAB = 180° - 80°

→ ∠CAB = 100°

now,

→ ∠DAE = ∠CAB { Vertically opposite angles. }

→ ∠DAE = 100°

then,

→ ∠FEA + ∠DAE = 180° { since L1 || L2 , interior angles on the same side of transversal are supplementry. }

→ x + 100° = 180°

→ x = 180° - 100°

→ x = 80° (Ans.)

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Answer 2

Solution :

In AABC we have,

→ 45° +35° + <CAB = 180° { By angle sum

property. }

→80° + angle CAB = 180°

→angle CAB = 180° - 80°

→ angle CAB = 100°

now,

→angle DAE = angle AB (Vertically opposite

angles. }

→ ZDAE = 100°

then,

→ angle FEA + angle AE = 180° { since L1 || L2, interior angles on the same side of

transversal are supplementry. }

→x +100° = 180°

→x = 180° - 100°

→ x = 80° (Ans.)