Question

please tell me the answer of this question ​

Answer 1

Answer:

angle D is 56°(alternate interior angle)

Answer 2

Given:

• AB || CD
• $\angle$ DAB = 56°
• $\angle$ APC = 100°

Find:

• $\angle$ PCD
• $\angle$ CPD

Solution:

we, know that

$\underline{\boxed{\sf\angle APC + \angle CPD = {180}^{ \circ} }} \quad \big\lgroup{\sf Linear \: Pair} \big\rgroup$

where,

• $\angle$ APC = 100°

So,

$\sf \dashrightarrow\angle APC + \angle CPD = {180}^{ \circ}$

$\sf \dashrightarrow {100}^{ \circ} + \angle CPD = {180}^{ \circ}$

$\sf \dashrightarrow\angle CPD = {180}^{ \circ} - {100}^{ \circ}$

$\sf \dashrightarrow\angle CPD = {80}^{ \circ}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \small{ \therefore\underline{\sf \angle CPD = {80}^{ \circ}}}$

____________________________

$\underline{\boxed{\sf \angle DAB = \angle ADC}} \quad \big\lgroup{\sf Alternate \: Angles} \big\rgroup$

where,

• $\sf\angle DAB$ = 56°

So,

$\sf : \to\angle DAB = \angle ADC$

$\sf : \to {56}^{ \circ} = \angle ADC$

$\sf : \to \angle ADC = {56}^{ \circ} \huge{......}1$

Now,

In ∆ PCD

$\underline{\boxed{\sf\angle PDC + \angle PCD + \angle CPD = 180^{\circ}}} \quad \big\lgroup{\sf Angle\: Sum \: Property} \big\rgroup$

where,

• $\angle$ CPD = 80°
• $\angle$ ADC = $\angle$ PDC = 56° [using eq.1]

So,

$\mapsto\sf\angle PDC + \angle PCD + \angle CPD = 180^{\circ}$

$\mapsto\sf 56^{\circ} + \angle PCD + 80^{\circ} = 180^{\circ}$

$\mapsto\sf \angle PCD + 136^{\circ} = 180^{\circ}$

$\mapsto\sf \angle PCD = 180^{\circ} - 136^{\circ}$

$\mapsto\sf \angle PCD = 44^{\circ}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \small{ \therefore\underline{\sf \angle PCD = {44}^{ \circ}}}$