Question

\_dcb=30 please solve this problem kl mera exam h ​

Answer 1

Question:

In the given figure , find Angle DCB, if AE || CD .

Answer:

30°

Note :

• Angle sum property of a triangle : The sum of all the three angles of a triangle is equal to 180° .

• Exterior angle property of a triangle : The exterior angle of a triangle is equal to the sum of two opposite interior angles .

• When two parallel lines are cut by a transversal then ;

• Corresponding angles are equal.
• Alternate interior angles are equal.
• Alternate exterior angles are equal.
• Consecutive interior angles are supplementary.

• Transversal line : It is a line which cut two other lines at two distinct points on the same plane.

• Complementary angles : Two angles are said to be complementary, if their sum is 90° .

• Supplementary angles : Two angles are said to be supplementary, if their sum is 180° .

Solution:

Given:

AE || CD

Angle BAE = 50°

Angle ABC = 20°

To find:

Angle DCB = ?

Construction:

Produce the line EA such that it intersect with the line PQ at the point L .

Now,

Let the Angle ALB = x.

Also,

For the ∆ABL ,

Angle BAE is the exterior angle and the two opposite interior angles corresponding to it are Angle ABL and Angle ALB .

Thus,

According to the exterior angle property of a triangle, we have ;

=> Angle BAE = Angle ABL + Angle ALB

=> 50° = 20° + x

=> x = 50° - 20°

=> x = 30°

Also,

It is given that, AE || CD.

Thus,

LE || CD

Since,

LE || CD and PQ is transversal ,

Thus,

=> Angle DCB = Angle ALB

{ corresponding angles are equal when a transversal cut two parallel lines }

=> Angle DCB = x

=> Angle DCB = 30°

Hence,

The required value of Angle DCB is 30° .

Answer 2

$\huge{\orange{\underline{\red{\mathfrak{GIVEN:-}}}}}$

1.<EAB=50°

2.<ABC=20°

$\huge{\orange{\underline{\red{\mathfrak{FIND:-}}}}}$

<DCB

$\huge{\orange{\underline{\red{\mathfrak{ANSWER:-}}}}}$

USING EXTERIOR ANGLE PROPERTY,

$\huge\bold\red\rightarrow$ $\huge\bold\red\rightarrow$<BAE=<ABX+<AXB

$\huge\bold\red\rightarrow$50°=20°+<AXB

$\huge\bold\red\rightarrow$<AXB=50°-20°

$\huge\bold\red\rightarrow$<AXB=30°

NOW,

$\huge\bold\red\rightarrow$APPLYING CORRESPONDING ANGLE THEOREM,

$\huge\bold\red\rightarrow$AC|| DC, XC TRANSVERSAL

$\huge\bold\red\rightarrow$.°. <AXB = <DCB

$\huge\bold\red\rightarrow$.°.<DCB=30°

$\huge{\orange{\underline{\red{\mathfrak{THANKS.}}}}}$