Question

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Answer 1

Solution :-

We need to find ,

• x = ?
• y = ?

By observing the given figure ,

♦ ABCD is a parallelogram with,

• Diagonal AC
• ∠B = 120°
• ∠CAD = 40°
• ∠CDA = x°
• Two congruent triangles ABC & CDA

♦ Right ∆CED with ,

• ∠DCE = y
• ∠E = 90°

Parallelogram ABCD :-

As we know that ,

Opposite angles of a parallelogram are always equal .

Here , opposite angles are ,

⇒ ∠B = ∠C

⇒ ∠C = 120° = x

Hence , x = 120° .

__________________

∆CED :-

Using Exterior angle property ,

• Exterior angle property :- Exterior angle equal to sum of interior angles .

• x = y + ∠E

Substituting know values ,

⇒ 120° = y + 90°

⇒ y = 120° - 90°

⇒ y = 30°

__________________

Hence , x = 120° & y = 30°

Answer 2

Given:

✪ABCD is an paralleogram where angle x = 120°

Find:

✪Measure of y

Solution:

Here,

$\sf \to \angle CDA + \angle CDE = {180}^{ \circ} \qquad \{ Linear Pair \}$

$\sf \to x+ \angle CDE = {180}^{ \circ}$

where, x = 120°

$\sf \to {120}^{ \circ}+ \angle CDE = {180}^{ \circ}$

$\sf \to \angle CDE = {180}^{ \circ} - {120}^{ \circ}$

$\boxed{ \sf \angle CDE = {60}^{ \circ}}$

$\rule{900}{2}$

$\sf \longrightarrow \angle CED + \angle CDE + \angle ECD= {180}^{ \circ} \qquad \{ angle \: sum \: property \}$

$\sf \longrightarrow {90}^{ \circ} + {60}^{ \circ} + y= {180}^{ \circ}$

$\sf \longrightarrow {150}^{ \circ} + y= {180}^{ \circ}$

$\sf \longrightarrow y= {180}^{ \circ} - {150}^{ \circ}$

$\underline{ \boxed{ \sf y= {30}^{ \circ}}}$

$\rule{900}{2}$

Hence, value of y will be 30°