Question

in the adjoining figure if AB is perpendicular to AD ,AB || CD , angle CBD = 30° and BCE =70° BAD =90° find the angles ABD and ADB ​

Answer 1

Answer:

• ∠ABD = 50°
• ∠ ADB = 40°

Step-by-step explanation:

GiveN :

• AB ⊥ AD
• AB||CD
• ∠CBD = 30°
• ∠BCE = 70°
• ∠BAD = 90°
• ∠ADC = 90°

To finD :

• ∠ABD
• ∠ ADB

Solution :

In ∆ BCD →

∠BCD + ∠BCE = 180° ------( linear pair )

→∠BCD + 70° = 180°

→∠BCD = 180° - 70°

→∠BCD = 110°

Now,

∠BDC + ∠BCD +∠DBC = 180° ----( angle sum property)

→∠BDC + 110° + 30° = 180°

→∠BDC + 140° = 180°

→∠BDC = 180° - 140°

→∠BDC = 40°

Now, In ∆ ABD →

[ We can write ∠ADC as ∠ADB + ∠BDC ]

∠ADB + ∠BDC = 90° -----( Given )

→∠ADB + 40° = 90°

→∠ADB = 90° - 40°

$\large{\bf{\green{\boxed{∠ADB = \:50°}}}}$

Now,

∠BAD + ∠ADB + ∠ABD = 180° ----(angle sum property)

→ 90° + 50° + ∠ABD = 180°

→ 140° + ∠ABD = 180°

→ ∠ABD = 180° - 140°

$\large{\bf{\green{\boxed{∠ABD=\:40°}}}}$

Answer 2

Answer:

Angle ADB = 50°

Angle ABD = 40°.

Step-by-step explanation:

In the figure -

Given -

• AB is perpendicular to AD.
• AB is parallel to CD.
• Angle CBD = 30°.
• Angle BCE = 70°.
• Angle BAD = 90°.

To find -

Angle ADB.

Angle ABD .

Solution -

Angle BCD + Angle BCE = 180° ( Linear pair).

then, angle BCD + 70° = 180°.

Angle BCD = 180° - 70° = 110°.

In ∆ BCD,

angle BDC + angle DBC + angle BCD = 180° ( angle sum property)

So, angle BCD + 30° + 110° = 180°.

angle BCD = 180°-140°=40°.

We know that -

AB is parallel to CD

then, BD is transversal

Angle ABD = Angle BDC = 40° ( by alternate interior angle).

Now, In ∆ ABD -

Angle ABD + Angle BAD + Angle ADB = 180° ( angle sum property).

40° + 90° + Angle ADB = 180° .

Angle ADB = 180° - 130° = 50° .

THEN,

Angle ABD = 40° .

Angle ADB= 50°.

my method is different .

I hope it's helpful for all of you.

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