In the given triangle AC = BC, find angles x and y, as shown in the figure.

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Solution

In $Δ A B C, A C = B C S o, a n g l e s o p p o s i t e t o e q u a l s i d e s w i l l b e e q u a l t o o . \Rightarrow ∠ A B C = ∠ B A C \Rightarrow ∠ A B C = x O p p o s i t e a n g l e s a r e e q u a l . s o, ∠ A C B = 92 ° T h e s u m o f a n g l e s o f a t r i a n g l e i s 180 ° s o, ∠ A B C + ∠ B A C + ∠ A C B = 180 ° \Rightarrow x + x + 92 ° = 180 ° \Rightarrow 2 x = 180 ° - 92 ° \Rightarrow 2 x = 88 ° \Rightarrow x = 44 ° N o w, w e h a v e t o f i n d y . U sin g e x t e r i o r a n g l e p r o p e r t y, y = ∠ B A C + ∠ A C B = 44 ° + 92 ° = 136 °$

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