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# ABCD is a rhombus and ABDE is a parallelogram.Given that EDC is a straight line and ∠AED = ${36^0}$ , find ∠BAD. $\left( {\text{a}} \right){\text{ 3}}{{\text{6}}^0} \\ \left( {\text{b}} \right){\text{ 7}}{{\text{2}}^0} \\ \left( {\text{c}} \right){\text{ 10}}{{\text{8}}^0} \\ \left( {\text{d}} \right){\text{ 12}}{{\text{0}}^0} \\ \$  Verified
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HINT- In order to solve such types of questions the key concept is that we should know the basic properties of rhombus and parallelogram along with the proper understanding of corresponding angles and alternate angles.

Corresponding angles - When two parallel lines (AE and BD) are crossed by another line EC (which is called the Transversal), the angles (∠BDC = ∠AED) in matching corners are called corresponding angles. ⇒ ∠BDC = ∠AED = ${36^0}$ (corresponding angles of AE and BD.)

⇒ We know that ABCD is a rhombus, so AB∥DC

Alternate angles-
Alternate angles are angles that are in opposite positions relative to a transversal (BD) intersecting two parallel lines (AB and DC). ⇒ ∠ABD = ∠BDC = ${36^0}$ (Alternate angles)

Isosceles triangle- The Isosceles Triangle Theorem states: If two sides of a triangle are equal, then the angles opposite those sides are equal.

⇒ ∠ADB = ∠ABD = ${36^0}$ (Base angles of isosceles, since AB =DC) Using formula $\left( {{\text{n - 2}}} \right) \times {180^0}$ , Here, number of side n=3
So sum of internal angle of triangle is ${180^0}$
⇒ ∠BAD + ∠ABD + ∠ADB = ${180^0}$
⇒ ∠BAD + ${36^0}$ + ${36^0}$= ${180^0}$
⇒ ∠BAD = ${180^0}$- ${36^0}$- ${36^0}$
∴ ∠BAD = ${108^0}$