Question

In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.

Answer 1

∠CED+∠BEC=180∘ (Linear Pair)

⇒ ∠CED+130∘=180∘

⇒ ∠CED+180°−130∘=50° ........(i)

Now, ∠ECD=200 .........(ii)

In ΔCED,

∠CED+∠ECD+∠CDE=1800 [Sum of all the angles of a triangle is 180∘]

⇒ 50°+20∘+∠CDE=180°   [Using (i) and (ii)]

⇒70°+∠CDE=180°

⇒ ∠CDE=180°−70°

⇒ ∠CDE=∠CDB=110° ........(iii)

Now ∠BAC=∠CDB=110∘  (angles in the same segment are equal)

Hence, ∠BAC=110°.