Q. 34.0( 4 Votes )

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.


Answer :

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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