In `triangleABC, AC=BC`, S is the circum-center and `angleASB=150^(@)` – InterviewSolution

1.	In `triangleABC, AC=BC`, S is the circum-center and `angleASB=150^(@)`. Find `angleCAB`. A. `55 (1/2^(@))`B. `52(1/2)^(@)`C. `90(1/2)^(@)`D. `35(1/2)^(@)`
Answer» Correct Answer - B Since, SA=SB=SC, S is the circum-center `angleASB=150^(@)` (given) `rArr angleACB=75^(@)`. Since, AC=BC, `angleCAB=angleCBA=(180^(@)-75^(@))/2` `=521/2^(@)`

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