Question

In the figure (3), EF ║GH, ∠EAB = 65° and ∠ACH = 100°. Find
(i) ∠ABC
(ii) ∠ACB
(iii) ∠CAF
(iv) ∠BAC

Answer 1

Answer:

i) ∠ABC = ?

Given

EF parallel to GH

therefore if two lines are parallel and a transversal intersects them,then their alternate interior angles are always equal,

that is

∠ABC=∠EAB

∠ABC=65°

ii) ∠ACB

Here from the figure,

∠ACB+∠ACH=180°(adjacent angles)

or it can be mentioned as linear pair/straight line

∠ACB+100=180

∠ACB=180-100

∠ACB=80°

iii) ∠CAF =?

Given

EF parallel to GH

therefore if two lines are parallel and a transversal intersects them,then their alternate interior angles are always equal,

that is

∠CAF=∠ACB

∠CAF= 80°

therefore

∠ACB=80°

(iv) ∠BAC

we know

∠BAC+∠ACB+∠ABC=180°(sum of all interior angles of a triangle are supplementary,that is 180°)

∠BAC+80+65=180

∠BAC+145=180

∠BAC=180-145

∠BAC=35°

Optional:-

∠EAB+∠BAC+∠CAF=180°(adjacent angles are supplementary)

65+35+80=180

180=180

L.H.S=R.H.S

hence verified

hope it helps