Question

in the given figure , find a) ANGLE ACD B)ANGLE AED​

Answer 1

• For finding Angle ACD

In ∆ABC,

By Angle Sum Property,

∠BAC + ∠ABC + ∠ACB = 180°

→ 15° + 50° + ∠ACB = 180°

→ 65° + ∠ACB = 180°

→ ∠ACB = 180° - 65°

→ ∠ACB = 115°

Now, by applying linear pair of angles,

∠ACB + ∠ACD = 180°

→ 115° + ∠ACD = 180°

→ ∠ACD = 180° - 115°

→ ∠ACD = 65°

OR

By Applying Exterior Angle Property,

∠ACD = ∠BAC + ∠ABC

→ ∠ACD = 15° + 50°

→ ∠ACD = 65°

• For finding ∠AED,

By applying Exterior Angle Property,

(or you can apply Angle Sum Property and Linear pair)

∠AED = ∠ACD + ∠EDC

→ ∠AED = 65° + 40°

→ ∠AED = 105°

Additional information:

• Angle Sum property states that Sum of all angles in a triangle is always 180°
• Exterior angle property states that the exterior angle is equal to the sum of opposite 2 interior angles.
• Linear pair of angles states that sum of given angles on a straight line is always 180°