We have ABIIED and BC||EF
(i) BC is transversal
∠DGC = ∠ABC [corresponding angles]
But ∠ABC = 70°
∴ ∠DGC = 70°
(ii) ED is a transversal to BC|lEF
∴ ∠DEF = ∠DGC [corresponding]
∠DGC = 70°
∠DEF = 70°

LEAB = 60° and LACD = 105°
Determine
(i) LCAF and
(ii) LBAC

(i) Since EF II GH and AC is a transversal
⇒ LCAF + LACH = 180
⇒ LCAF + 105° = 180°
=75°
(ii) ∴ EF II GH and AC is transversal.
∴ LEAC = LACH
⇒ LBAC = 105°
⇒ LBAC + LBAB = 105°
⇒ LBAC + 60° = 105°
⇒ LBAC = 105°- 60°
= 45°
∴ LCAF = 75° and LBAC = 45°

No, two lines cannot intersect in more than one point.

(i) Since the angles are linear pair, \(\angle\)ACD + \(\angle\)BCD = 180°
123° + \(\angle\)BCD = 180°
Subtracting 123° on both sides
123° + \(\angle\)BCD – 123° = 180° – 123°
\(\angle\)BCD = 57°
(ii) Since the angles are linear pair, \(\angle\)LNO + \(\angle\)MNO = 180°
46° + \(\angle\)MNO = 180°
Subtracting 46° on both sides
46° + \(\angle\)MNO – 46° = 180° – 46°
\(\angle\)MNO = 134°

We know that \(\angle\)POR = \(\angle\)POQ + \(\angle\)QOR
62° = 23° + \(\angle\)QOR
Subtracting 23° on both sides
62° – 23° = 23° + \(\angle\)QOR – 23°
\(\angle\)QOR = 39°

∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° - 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles
∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC = 130°
∠BOD = 130°

∠BAD = ∠BAE + ∠EAD
= 40° + 30° = 70°.
and LCDA = 70°
∠BAD = ∠CDA
But they form a pair of alternate angles
⇒ AB || CD
Also ∠BAE + ∠AEF = 40° + 140° = 180° -------(1)
But they form a pair of interior opposite angles.
⇒ AB II EF
From (1) and (2), we get ---(2)
AB || CD || EF
⇒ CD II EF

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

(i) The corresponding angles to b° are d° and 68°.
(ii) The measure of b° is 68° (Because b° corresponds to 68°)
(iii) The angles having measure 68° are b°, d°, g°, i° and k°.
(iv) The angles having measure 112° are a°, c°, e°, f°, h° and j°.

x = 55° vertically opposite angles
y+ 55° = 180°
y = 180°- 55°
y = 125°
z = 125°

Yes, because the sum of two right angles is 180° and form a linear pair.

No, the sum of the measures of two obtuse angles is more than 180°.

Yes, In the figure
∠AOB and ∠BOC are adjacent angles.
Also ∠AOB + ∠BOC = 180°
∴ ∠AOB and ∠BOC are supplementary

From this figure
<CDE = <EAB
(Alternate interior angles)
x^o = 48^o
<ABE = <DCE
(Alternate interior angles)
y^o = 60^o
We know that z^o = y^o + 48^o
= 60^o + 48^o
= 108^o