Question

Calculate the value of x in each of the following figures.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer 1

ANSWER:
(i)
Side AC of triangle ABC is produced to E .
∴∠EAB=∠B+∠C⇒110°=x+∠C ...(i)
Also,
∠ACD+∠ACB=180°

⇒120°+∠ACB=180°⇒∠ACB=60°⇒∠C=60°
Substituting the value of ∠C in i, we get x=50

(ii)
From ∆ABC we have:
∠A+∠B+∠C=180°

⇒30°+40°+∠C=180°⇒∠C=110°⇒∠ACB=110°
Also,
∠ECB+∠ECD=180° ( linear pair)

⇒110°+∠ECD=180°⇒∠ECD=70°Now, in ∆ECD,∴∠AED=∠ECD+∠EDC
⇒x=70°+50°⇒x=120°

(iii)
∠ACB+∠ACD=180° (linear pair)

⇒∠ACB+115°=180°⇒∠ACB=65°
Also,
∠EAF=∠BAC Vertically opposite angles

⇒∠BAC=60°∴∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle)

⇒60°+x+65°=180°⇒x=55°
(iv)
∠BAE=∠CDE (Alternate angles)

⇒∠CDE=60° (∴∠ECD+∠CDE+∠CED=180°)
⇒x=75°
Sum of the angles of a triangle⇒45°+60°+x=180°

(v)
From ∆ABC, we have:
∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle)

⇒40°+∠ABC+90°=180°⇒∠ABC=50°
Also, from ∆EBD, we have:
∠BED+∠EBD+∠BDE=180° (Sum of the angles of a triangle)

⇒100°+50°+x=180° ∵∠ABC=∠EBD⇒x=30°

(vi)
From ∆ABE, we have:
∠BAE+∠ABE+∠AEB=180° (Sum of the angles of a triangle)

⇒75°+65°+∠AEB=180°⇒∠AEB=40° ∴∠AEB=∠CED (Vertically opposite angles)

⇒∠CED=40°
Also, From ∆CDE, we have
∠ECD+∠CDE+∠CED=180° (Sum of the angles of a triangle)

⇒110°+x+40°=180°⇒x=30°