(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)

Solution:

(a) In ∆PQR,

∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58°

∠RPQ = 90° – ∠PQR = 90° – 58° = 32°

SR || PQ (given)

∠SRP = ∠RPQ = 32^o (Alternate angles)

Now PRST is a cyclic quadrilateral,

∠STP + ∠SRP = 180^o

∠STP = 180^o – 32^o = 148^o

(b) In the given figure,

∠ACE 43^o and ∠CAF = 62⁰

Now, in ∆AEC

∠ACE + ∠CAE + ∠AEC = 180^o

43^o + 62^o + ∠AEC = 180^o

105^o + ∠AEC = 180^o

∠AEC = 180^o – 105⁰ = 75^o

But ∠ABD + ∠AED = 180⁰

(sum of opposite angles of acyclic quadrilateral)

and ∠AED = ∠AEC

a + 75^o = 180^o

a = 180^o – 75^o – 105^o

but ∠EDF = ∠BAE

(Angles in the alternate segment)