# In Fig. 9.30, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC.

Given,

ACD = 105o

EAF = 45o

EAF = BAC (Vertically opposite angle)

BAC = 45o

ACD + ACB = 180o (Linear pair)

105o + ACB = 180o

ACB = 180o – 105o

= 75o

In

BAC + ABC + ACB = 180o

45o + ABC + 75o = 180o

ABC = 180o – 120o

= 60o

Thus, all three angles of a triangle are 45o, 60o and 75o.

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