Question

In the given figure, EF || GH, angle EAB = 65°
and angleACH = 100º. Determine (i) angle ABC (ii)
angle ACB (iii) angle BAC (iv) angleCAF.​

Answer 1

Given : EF || GH angle Eab =65° and angle ACH=100°

To Find : (i) angle ABC (ii)  angle ACB (iii) angle BAC (iv) angleCAF.​

Solution:

∠ACB & ∠ACH form a linear pair

=> ∠ACB + ∠ACH = 180°

∠ACH = 100°

=> ∠ACB + 100° = 180°

=>  ∠ACB  = 80°

EF || GH  & AB is transversal

∠ABC  = ∠EAB    alternate interior angle

∠EAB    = 65°

∠ABC =  65°

∠CAF = ∠ACB   alternate interior angle

=> ∠CAF = 80°

∠ABC + ∠ACB + ∠BAC = 180°  Sum of angles of triangle

=> 65° + 80° + ∠BAC = 180°

=> ∠BAC =35°

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Answer 2

Solution:

∠ACB & ∠ACH form a linear pair

=> ∠ACB + ∠ACH = 180°

∠ACH = 100°

=> ∠ACB + 100° = 180°

=>  ∠ACB  = 80°

EF || GH  & AB is transversal

∠ABC  = ∠EAB    alternate interior angle

∠EAB    = 65°

∠ABC =  65°

∠CAF = ∠ACB   alternate interior angle

=> ∠CAF = 80°

∠ABC + ∠ACB + ∠BAC = 180°  Sum of angles of triangle

=> 65° + 80° + ∠BAC = 180°

=> ∠BAC =35°