Question

Given that DAB and EAC are straight lines find the value of unknown.​

Answer 1

Given:

→ Angle CAB = 110°

→ CD = CA

→ ∆CDA and DA is produced to B

To find:

→ x°

Solution:

Angle DAB = 180° (straight line)

Angle DAC + Angle CAB = 180° ( Angle DAB = Angle DAC + Angle CAB)

→ We know that,

Angle CAB = 110°

Angle DAC + 110° = 180° (Measure of Angle CAB = 110°)

Angle DAC = 180° - 110°

→ Angle DAC = 70° ←

→ We know that,

CD = CA

Angles opposite to equal sides are equal

CD = CA

Angle CAD = Angle CDA

→ We know that,

Angle CAD = 70°

→ Therefore, Angle CDA = 70°

→ Now,

Angle CDA + Angle DAC + Angle ACD = 180° (Angle - Sum property of a ∆)

70° + 70° + x° = 180° ( Angle CDA = Angle DAC = 180°, Angle ACD = x°)

140° + x° = 180°

x° = 180° - 140°

→ x° = 40° ←

Conclusion:

Therefore, Angle ACD = x° = 40°

Extra Information:

→ Angle - Sum property or Sum of all angles in a ∆ is 180°

→ Angles opposite to equal sides are equal.

→ If 2 sides in a ∆ are equal then, that particular ∆ is called as Isosceles ∆.

Answer 2

Given:

→ Angle CAB = 110°

→ CD = CA

→ ∆CDA and DA is produced to B

To find:

→ x°

Solution:

Angle DAB = 180° (straight line)

Angle DAC + Angle CAB = 180° ( Angle DAB = Angle DAC + Angle CAB)

→ We know that,

Angle CAB = 110°

Angle DAC + 110° = 180° (Measure of Angle CAB = 110°)

Angle DAC = 180° - 110°

→ Angle DAC = 70° ←

→ We know that,

CD = CA

Angles opposite to equal sides are equal

CD = CA

Angle CAD = Angle CDA

→ We know that,

Angle CAD = 70°

→ Therefore, Angle CDA = 70°

→ Now,

Angle CDA + Angle DAC + Angle ACD = 180° (Angle - Sum property of a ∆)

70° + 70° + x° = 180° ( Angle CDA = Angle DAC = 180°, Angle ACD = x°)

140° + x° = 180°

x° = 180° - 140°

→ x° = 40° ←

Conclusion:

Therefore, Angle ACD = x° = 40°

Extra Information:

→ Angle - Sum property or Sum of all angles in a ∆ is 180°

→ Angles opposite to equal sides are equal.

→ If 2 sides in a ∆ are equal then, that particular ∆ is called as Isosceles ∆.