Question

2. In the given figure, ABCD is a
parallelogram, in which DAB = 75 and
DBC = 60°. Calculate CDB and ADB?
​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\angle ADB=60\degree}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline\bold{Given : }} \\ \tt: \implies \angle DAB = 75 \degree \\ \\ \tt: \implies \angle CBD = 60 \degree \\ \\ \red{\underline\bold{To \:Find : }} \\ \tt: \implies \angle ADB= ?$

• According to given question :

$\circ \: AB || \: CD\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \angle CBD = \angle ADB \: \: \: \: (Alternate \: angle) \\ \\ \tt: \implies \angle 60 \degree = \angle ADB \\ \\ \green{\tt: \implies \angle ADB = 60 \degree} \\ \\ \bold{Alternate \: method} \\ \tt : \implies \angle DAB + \angle ABC = 180 \degree \: \: \: \: (Adjacent \: angle) \\ \\ \tt: \implies 75 \degree + \angle ABD+ \angle CBD = 180 \degree \\ \\ \tt: \implies \angle ABD + 60 \degree = 180 \degree - 75 \degree \\ \\ \tt: \implies \angle ABD = 105 \degree - 60 \degree \\ \\ \tt: \implies \angle ABD = 45 \degree \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \angle ABD + \angle \: DAB + \angle ADB = 180 \degree \: \: \: \: (sum \:of \: all \: angle \: of \: triangle \: is \: 180 \degree) \\ \\ \tt: \implies 45 \degree + 75 \degree + \angle ADB = 180 \degree \\ \\ \tt: \implies \angle ADB= 180 \degree - 120 \degree \\ \\ \green{\tt: \implies \angle ADB= 60 \degree}$

Answer 2

Answer:

★ ∠CDB = 45° and ∠ADB =60° ★

Step-by-step explanation:

Given:

• ∠DAB is 75° and ∠DBC is 60°

To Find:

• ∠CDB and ∠ADB

Solution: As we know that the opposite sides of a Parallelogram are Parallel to each other and the sum of all angles of a Parallelogram is 180°

∴ ∠ADB = ∠CBD = 60° [ Alternate interior angles ]

†Now, In ∆ADB, By Angle Sum Property †

→ ∠DAB + ∠ADB + ∠ABD = 180°

→ 75° + 60° + ∠ABD = 180°

→ 135° + ∠ABD = 180°

→ ∠ABD = 180 – 135

→ ∠ABD = 45°

∴ ∠CDB = ∠ABD = 45° [ Alternate Interior Angles ]

Hence, ∠CDB = 45° and ∠ADB = 60°