In a triangle ABC AC = BC. Angle C is 116 degrees. Find the outside corner of the CBD.

If AC = BC, then <A = <B;
If <C = 116 degrees, then <A + <B = 180 – 116 = 64;
Since <A = <B, it means <A = <B = 64/2 = 32;
Now we can find the outer corner of the CBD.
<CBD = 180 – <B = 180 – 32 = 148.
Answer: The outside angle of the CBD is 148 degrees.



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